What is the relationship between the PSD of a continuous signal and the PSD of its periodically sampled one?
If you sample a finite-power continuous-time random process $x(t)$ you get a discrete-time random process $y_k$. If $x(t)$ is wide-sense stationary (WSS) you get for the autocorrelation function of $y_k$

$$R_y(k,l)=E\{X(kT)X^*(lT)\}=R_x((k-l)T)=R_x(mT)$$

where $T$ is the sampling period. Obviously, $y_k$ is also WSS (it only depends on the difference $m=k-l$), and its autocorrelation function is a sampled version of the autocorrelation $R_x(\tau)$ of the original continuous-time process. Since the power spectrum is the Fourier transform of the autocorrelation function, the relation between the two power spectra is the same as the relation between the spectrum of a continuous signal and its sampled version:

$$S_y(e^{j\omega T})=\frac{1}{T}\sum_{k=-\infty}^{\infty}S_x(\omega-2\pi k/T)\tag{1}$$

Since the anti-aliasing filter has a cut-off frequency of $\pi/T$ it is sufficient to consider the interval $-\pi/T\le\omega&lt;\pi/T$ (i.e. just consider the term $k=0$ in (1)):

$$S_y(e^{j\omega T})=\frac{1}{T}S_x(\omega)\tag{2}$$

This equation is equivalent to Eq. (10.65) in your question. Just note that I've used $\omega$ in a different way (analog frequency in radians), whereas Oppenheim uses $\omega$ as the normalized angular frequency. So with $\omega'=\omega T$ you finally get

$$S_y(e^{j\omega'})=\frac{1}{T}S_x\left(\frac{\omega'}{T}\right)$$