This is the equation I get:
\begin{equation}
X(\omega) = \sum ^{\infty}_{k=-\infty} X_a\left \lbrace \frac{\omega - 2\pi k}{T_s}\right \rbrace
\end{equation}
Here I used the convention of lower case to represent normalized frequencies, determined by dividing the frequency in cycles/sec or radians/sec by the sampling rate to get cycles/sample or radians/sample.
The above relationship assumes using $X_a(\Omega)$ with $\Omega$ as the angular frequency in radians/sec. It appears the equation posted by the OP is using $X_a(F)$ with $F$ in cycles/sec or Hz, and scaling the output by the sampling rate. If that is what was intended, then the equation appears correct for that purpose- but that should be specified.
My assumption is the equation is describing the result of sampling in time (at sampling rate $F_s = 1/T_s$) creating periodicity in frequency, here over index $k$.
Similarly, as a function of $\Omega$ in radians/sample we would get:
$$X(\Omega) = \sum ^{\infty}_{k=-\infty} X_a\left \lbrace {\Omega - \frac{2\pi k}{T_s}\right \rbrace $$
The difference between this and the OP’s answer below is I am assuming “sampling” is done as a product with an impulse train while the alternate approach uses discrete time approximation of integration in that each “sample” is the area under the curve over interval $T_s$ and dividing by $T_s$ provides the average amplitude over that time interval.
After sampling, we expect the analog spectrum from $F=0$ (in Hz) to $F=F_s$ (in Hz) to periodically repeat as $f$ extends to $\pm \infty$.
$\omega = 2\pi f$ is the continuous frequency domain in units of radians/sample and here extending periodically to $\pm \infty$. $f= F/F_s$ and is the normalized frequency domain in units of cycles/sample.
$X_a(\Omega)$ is the analog spectrum, with $\omega= 2\pi F$ in units of radians/sec. (And bandlimited to zero outside the range from $-2\pi F_s/2$ to $+2\pi F_s/2$ if we wish to avoid aliasing)
$X(\omega)$ is the resulting continuous and periodic spectrum of the discrete time signal that is periodic over any interval from $(k-1) 2\pi$ to $k 2 \pi$ for any $k$. (Integer $k$ used in the original formula), with $\omega$ in units of radians/sample.
With that, we are mapping the normalized frequency units of $\omega$ as radians/sample used for the discrete time spectrum $X(\omega)$ to the frequency units of $\Omega$ as radians/second and capturing the additive effects of the aliasing and periodicity over the $kf_s$ intervals in $\Omega$.
This is given by
$$\omega \rightarrow \Omega$$
$$\omega \rightarrow \frac{\omega-2\pi}{T_s}$$
With that we get the expected periodic spectrum for $X(\omega)$ as we index through each $k$.
This would result in the resulting spectrum having the same scale as the original spectrum at all the original frequency locations. If we were to multiply this result by $1/Ts$ then that would scale the result by the sampling rate, which alone does not make sense to me as an artifact of the sampling process specifically: Sampling in time is equivalent to multiplying the time domain waveform with a stream of impulses spaced by the sampling rate, and each with an area of 1. This product in time is a convolution in frequency resulting in a copy of the original spectrum with the same scale centered around the frequency of each of those impulses (which are spaced by the sampling rate), with no further scaling.