In [1, p. 62 exercise 3.13] it is requested to prove that the PSD of a real WSS random process is a real even function of frequency. Solution: The PSD of a WSS process is given by the Fourier transform of its autocorrelation function:
P_{xx}(f)=\sum\limits_{k=-\infty}^{\infty}r_{xx}[k]e^{-j2\pi f k}. (1)

We note that for a real process x[n], r_{xx}[k] will be also real thus we obtain the relations:
r^{\ast}_{xx}[k]=r_{xx}[-k]=r_{xx}[k]. (2)

First let us prove that for a real WSS this function is even. Having the previous relations in mind:
P_{xx}(-f)=\sum\limits_{k=-\infty}^{\infty}r_{xx}[k]e^{j2\pi f k}
=\sum\limits_{k=-\infty}^{\infty}r_{xx}[-k]e^{j2\pi f k}

Let u=-k then
P_{xx}(-f)=\sum\limits_{u=-\infty}^{\infty}r_{xx}[u]e^{-j2\pi f u}
=P_{xx}(f) (3)

Thus we have shown that the PSD is symmetric P_{xx}(f)=P_{xx}(-f). The next step is to show that the PSD is also real. We know that the real part of a complex function is given as one half of the sum of the complex function with its complex conjugate form, that is Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P^{\ast}_{xx}(f)\right). Let’s determine the complex conjugate of the PSD:
P_{xx}^{\ast}(f)=\sum\limits_{k=-\infty}^{\infty}r^{\ast}_{xx}[k]e^{j2\pi f k}
=\sum\limits_{k=-\infty}^{\infty}r_{xx}[-k]e^{j2\pi f k}=P_{xx}(-f)=P_{xx}(f) (4)

Again relation (2) was used in order to obtain the last result. The real part of the PSD is thus given by
Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P^{\ast}_{xx}(f)\right)
=Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P_{xx}(f)\right)
=P_{xx}(f),

because of (4). So the real part of the PSD P_{xx}(f) is the PSD itself, or stating it in another way: the PSD of a WSS process is real. Thus together with (3) we have shown that the PSD of a real WSS random process is a real even function of frequency. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.