Let's start with your equation.

$$ P_k\left( e^{j\Omega} \right) = \frac{1}{L} e^{jk\Omega/L} \sum_{p=0}^{L-1} e^{-2 \pi j k p / L} H\left( e^{j(\Omega-2\pi p)/L} \right) $$

Do a litte rearranging.

$$ P_k\left( e^{j\Omega} \right) = e^{j\Omega\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} H\left( e^{j\Omega\frac{1}{L} } e^{-j2\pi \frac{p}{L}} \right) \left( e^{j2\pi/L} \right)^{-kp} \right] $$

It can be made simpler to understand by a few variable substitutions.

$$ y = e^{j\Omega} $$

$y$ is a complex value on the complex unit circle at $\Omega$ radians around the circumference.

$$ h[p] = H\left( y^{\frac{1}{L}} e^{-j2\pi \frac{p}{L}} \right) $$

$h[]$ is a sequence of values sampled from the $H()$ function. $y^{\frac{1}{L}}$ is a complex value on the unit circle at $1/L$th the angle of $y$. The $e^{-j2\pi \frac{p}{L}}$ is a clockwise walk around the unit circle in $L$ even steps. Conceptually:

$$ e^{-j2\pi \frac{p}{L}} = \left( e^{j2\pi } \right)^{-\frac{p}{L}} = 1^{-\frac{p}{L}} $$

Therefore $h[]$ has a period of $L$. If $p$ were to go below zero or above $L-1$, it would be a repeat pattern.

With these substitutions, your equation becomes this:

$$ P_k\left( y \right) = y^{\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} h[p] \left( e^{j2\pi/L} \right)^{-pk} \right] $$

The part in brackets is the definition of the DFT.

$$ X[k] = \sum_{n=0}^{N-1} x[n] \left(e^{i2\pi/N}\right)^{-nk} $$

It's arranged the same way as in my answer here Fractional powers of complex numbers (DSPrelated computation)

The $\frac{1}{L}$ is a normalizing factor making the magnitudes of the DFT calculation independent of $L$.

$ y^{\frac{k}{L}} $ is a complex value $ k/L $ of the way to $y$ along the unit circle which means the results of the DFT bin are rotated that far.

So, a long answer to your question. The purpose of the second exponential $ e^{-2\pi j k p/L} $ is to execute the DFT transform. It is part of the definition and the reason it works the way it does. Lots of stuff available on that if you are unfamiliar.

Let's start with your equation.

$$ P_k\left( e^{j\Omega} \right) = \frac{1}{L} e^{jk\Omega/L} \sum_{p=0}^{L-1} e^{-2 \pi j k p / L} H\left( e^{j(\Omega-2\pi p)/L} \right) $$

Do a little rearranging.

$$ P_k\left( e^{j\Omega} \right) = e^{j\Omega\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} H\left( e^{j\Omega\frac{1}{L} } e^{-j2\pi \frac{p}{L}} \right) \left( e^{j2\pi/L} \right)^{-kp} \right] $$

It can be made simpler to understand by a few variable substitutions.

$$ y = e^{j\Omega} $$

$y$ is a complex value on the complex unit circle at $\Omega$ radians around the circumference.

$$ h[p] = H\left( y^{\frac{1}{L}} e^{-j2\pi \frac{p}{L}} \right) $$

$h[]$ is a sequence of values sampled from the $H()$ function. $y^{\frac{1}{L}}$ is a complex value on the unit circle at $1/L$th the angle of $y$. The $e^{-j2\pi \frac{p}{L}}$ is a clockwise walk around the unit circle in $L$ even steps. Conceptually:

$$ e^{-j2\pi \frac{p}{L}} = \left( e^{j2\pi } \right)^{-\frac{p}{L}} = 1^{-\frac{p}{L}} $$

Therefore $h[]$ has a period of $L$. If $p$ were to go below zero or above $L-1$, it would be a repeat pattern.

With these substitutions, your equation becomes this:

$$ P_k\left( y \right) = y^{\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} h[p] \left( e^{j2\pi/L} \right)^{-pk} \right] $$

The part in brackets is the definition of the DFT.

$$ X[k] = \sum_{n=0}^{N-1} x[n] \left(e^{i2\pi/N}\right)^{-nk} $$

It's arranged the same way as in my answer here Fractional powers of complex numbers (DSPrelated computation)

The $\frac{1}{L}$ is a normalizing factor making the magnitudes of the DFT calculation independent of $L$.

$ y^{\frac{k}{L}} $ is a complex value $ k/L $ of the way to $y$ along the unit circle which means the results of the DFT bin are rotated that far.

So, a long answer to your question. The purpose of the second exponential $ e^{-2\pi j k p/L} $ is to execute the DFT transform. It is part of the definition and the reason it works the way it does. Lots of stuff available on that if you are unfamiliar.

Let's start with your equation.

$$ P_k\left( e^{j\Omega} \right) = \frac{1}{L} e^{jk\Omega/L} \sum_{p=0}^{L-1} e^{-2 \pi j k p / L} H\left( e^{j(\Omega-2\pi p)/L} \right) $$

Do a litte rearranging.

$$ P_k\left( e^{j\Omega} \right) = e^{j\Omega\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} H\left( e^{j\Omega\frac{1}{L} } e^{-j2\pi \frac{p}{L}} \right) \left( e^{j2\pi/L} \right)^{-kp} \right] $$

It can be made simpler to understand by a few variable substitutions.

$$ y = e^{j\Omega} $$

$y$ is a complex value on the complex unit circle at $\Omega$ radians around the circumference.

$$ h[p] = H\left( y^{\frac{1}{L}} e^{-j2\pi \frac{p}{L}} \right) $$

$h[]$ is a sequence of values sampled from the $H()$ function. $y^{\frac{1}{L}}$ is a complex value on the unit circle at $1/L$th the angle of $y$. The $e^{-j2\pi \frac{p}{L}}$ is a clockwise walk around the unit circle in $L$ even steps. Conceptually:

$$ e^{-j2\pi \frac{p}{L}} = \left( e^{j2\pi } \right)^{-\frac{p}{L}} = 1^{-\frac{p}{L}} $$

Therefore $h[]$ has a period of $L$. If $p$ were to go below zero or above $L-1$, it would be a repeat pattern.

With these substitutions, your equation becomes this:

$$ P_k\left( y \right) = y^{\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} h[p] \left( e^{j2\pi/L} \right)^{-pk} \right] $$

The part in brackets is the definition of the DFT.

$$ X[k] = \sum_{n=0}^{N-1} x[n] \left(e^{i2\pi/N}\right)^{-nk} $$

It's arranged the same way as in my answer here Fractional powers of complex numbers (DSPrelated computation)

The $\frac{1}{L}$ is a normalizing factor making the magnitudes of the DFT calculation independent of $L$.

$ y^{\frac{k}{L}} $ is a complex value $ k/L $ of the way to $y$ along the unit circle.

So, a long answer to your question. The purpose of the second exponential $ e^{-2\pi j k p/L} $ is to execute the DFT transform. It is part of the definition and the reason it works the way it does. Lots of stuff available on that if you are unfamiliar.

Let's start with your equation.

$$ P_k\left( e^{j\Omega} \right) = \frac{1}{L} e^{jk\Omega/L} \sum_{p=0}^{L-1} e^{-2 \pi j k p / L} H\left( e^{j(\Omega-2\pi p)/L} \right) $$

Do a litte rearranging.

$$ P_k\left( e^{j\Omega} \right) = e^{j\Omega\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} H\left( e^{j\Omega\frac{1}{L} } e^{-j2\pi \frac{p}{L}} \right) \left( e^{j2\pi/L} \right)^{-kp} \right] $$

It can be made simpler to understand by a few variable substitutions.

$$ y = e^{j\Omega} $$

$y$ is a complex value on the complex unit circle at $\Omega$ radians around the circumference.

$$ h[p] = H\left( y^{\frac{1}{L}} e^{-j2\pi \frac{p}{L}} \right) $$

$h[]$ is a sequence of values sampled from the $H()$ function. $y^{\frac{1}{L}}$ is a complex value on the unit circle at $1/L$th the angle of $y$. The $e^{-j2\pi \frac{p}{L}}$ is a clockwise walk around the unit circle in $L$ even steps. Conceptually:

$$ e^{-j2\pi \frac{p}{L}} = \left( e^{j2\pi } \right)^{-\frac{p}{L}} = 1^{-\frac{p}{L}} $$

Therefore $h[]$ has a period of $L$. If $p$ were to go below zero or above $L-1$, it would be a repeat pattern.

With these substitutions, your equation becomes this:

$$ P_k\left( y \right) = y^{\frac{k}{L}} \frac{1}{L} \left[ \sum_{p=0}^{L-1} h[p] \left( e^{j2\pi/L} \right)^{-pk} \right] $$

The part in brackets is the definition of the DFT.

$$ X[k] = \sum_{n=0}^{N-1} x[n] \left(e^{i2\pi/N}\right)^{-nk} $$

It's arranged the same way as in my answer here Fractional powers of complex numbers (DSPrelated computation)

The $\frac{1}{L}$ is a normalizing factor making the magnitudes of the DFT calculation independent of $L$.

$ y^{\frac{k}{L}} $ is a complex value $ k/L $ of the way to $y$ along the unit circle which means the results of the DFT bin are rotated that far.

So, a long answer to your question. The purpose of the second exponential $ e^{-2\pi j k p/L} $ is to execute the DFT transform. It is part of the definition and the reason it works the way it does. Lots of stuff available on that if you are unfamiliar.