Good question, which I'll try and answer without complete mathematical rigorousness, but focus more on the intuition behind.
First, let's define the periodogram of a real discrete signal $x[n]$ as the Discrete Fourier Transform of its auto-correlation function:
$$\mathcal{F}\{x[n]*x[-n]\} = \big|X[k]\big|^2$$
Here, $n$ is the sample number, $*$ the convolution operator, $\mathcal{F}$ the Discrete Fourier Transform, $X$ the Discrete Fourier Transform (I know, both the transform and the result share the same name), $k$ the frequency bin number and $\big|\cdot\big|$ the absolute value operator. $\big|X[k]\big|$ is called the magnitude spectrum.
In plain english, the periodogram of a signal $x[n]$ is an estimate of the spectral power content of $x[n]$, in other words, an estimate of the power at each frequency that make up $x[n]$.
The question is, for two signals $x_1[n]\neq x_2[n]$, how can $\big|X_1[k]\big|^2 = \big|X_2[k]\big|^2$
This can be answered using a combination of intuition and mathematics using the definition for the Discrete Fourier Transform. Let's call $f_s$ the sampling frequency with which $x_1[n]$ and $x_2[n]$ were sampled.
As I'm sure you're aware, the Discrete Fourier Transform can decompose a time-domain signal into its individual frequency components.
Ask yourself, what frequencies make up a chirp signal $x_1[n]$? That one is easy to answer intuitively, since a chirp signal is a signal in which the frequency increases (or decreases) with time, starting at frequency $f_0$ and ending at $f_1$. Since our signal is sampled at $f_s$, let's conveniently set $f_0 = 0\,\text{Hz}$ and $f_1 = f_s/2\,\text{Hz}$: the chirp is made-up of all frequencies between $0$ and $f_s/2\,\text{Hz}$ (discarding frequency resolution considerations here). Assuming the amplitude remains constant and arbitrarily setting it to $1$, that would give you, without the need for complicated mathematics:
$$\big|X_1[k]\big|^2 = 1$$
Less intuitive question: what frequencies make up an impulse signal? Well, the answer is all of them. To show this, let's use the Discrete Fourier transform:
$$X[k] = \sum_{n=0}^{N-1}x[n]e^{-j2\pi kn/N}$$
and compute it for $x_2[n] = \delta[n-m] = \begin{cases}1 &n=m\\ 0 &n\neq m\end{cases}$
$$
X_2[k] = \sum_{n=0}^{N-1}\delta[n-m].e^{-j2\pi kn/N} = e^{-j2\pi k m/N}
$$
The periodogram does not care about phase, it is only interested in the (squared) magnitude of $X_2[k]$:
$$\big|X_2[k]\big|^2 = \big|e^{-j2\pi k m/N}\big|^2 = 1$$
So here you have it:
$$\big|X_1[k]\big|^2 = \big|X_2[k]\big|^2$$
