In [1, p. 62 exercise 3.15] it is requested to verify the ACF and PSD relationships given in [1, p. 53 eq. (3.50)] and [1, p. 54 eq. (3.51)].
Solution: Starting from the first relationship of [1, p. 53 eq. (3.50)] and the definition of the cross correlation function we can derive
r_{xy}[n,k]=E\{x^{\ast}[n]y[n+k]\}
=E\{x^{\ast}[n]\sum\limits_{l}x[n+k-l]h[l]\}
=\sum\limits_{l}E\{x^{\ast}[n]x[n+k-l]\}h[l]

Using the starting assumption that x[n] is WSS we can derive the final result, that the cross correlation is also independent of the observation instance n and depends only on the lag k:
r_{xy}[n,k]=r_{xy}[k]=\sum\limits_{l}r_{xx}[k-l]h[l]

Similar the second relation can be obtained by:
r_{yx}[n,k]=E\{y^{\ast}[n]x[n+k]\}
=E\{\sum\limits_{l}x^{\ast}[n-l]h^{\ast}[l]x[n+k]\}
=\sum\limits_{l}E\{x^{\ast}[n-l]x[n+k]\}h^{\ast}[l]
=\sum\limits_{l}r_{xx}[k+l]h^{\ast}[l]

and setting l^{\prime} = -l:
r_{yx}[n,k]=\sum\limits_{l^{\prime}}r_{xx}[k-l^{\prime}]h^{\ast}[-l^{\prime}]
=r_{xx}[k]\star h^{\ast}[-k] =r_{yx}[k]

The third equation can be derived by:
r_{yy}[n,k]=E\{y^{\ast}[n]y[n+k]\}
=E\{\sum\limits_{l}h^{\ast}[l]x^{\ast}[n-l]\sum\limits_{\lambda}h[\lambda]x[n+k-\lambda] \}
=\sum\limits_{l}\sum\limits_{\lambda}h^{\ast}[l]h[\lambda]E\{x^{\ast}[n-l]x[n+k-\lambda] \}
=\sum\limits_{l}h^{\ast}[l]\sum\limits_{\lambda} h[\lambda]r_{xx}[k+l-\lambda] (1)

By setting \rho[k+l]=\sum_{\lambda} h[\lambda]r_{xx}[k+l-\lambda] =h[k+l]\ast r_{xx}[k+l] we can rewrite (1) by
r_{yy}[n,k]=\sum\limits_{l}h^{\ast}[l] \rho[k+l]
=h^{\ast}(-k)\star \rho(k)
=h^{\ast}(-k)\star h[k] \ast r_{xx}[k] (2)

which proves the last equation of [1, p. 53 eq. (3.50)] . We can now proceed to prove the relations for [1, p. 54 eq. (3.51)]
P_{xy}(z)=\mathcal{Z}\{r_{xy}[k]\}
=\mathcal\{r_{xx}[k]\star h[k]\}
=H(z)R_{xx}(z) (3)

The second relation P_{yx}(z) can similar be proven by:
P_{yx}(z)=\mathcal{Z}\{r_{yx}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\star r_{xx}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\}R_{xx}(z)
=R_{xx}(z)\sum\limits_{k=-\infty}^{\infty}h^{\ast}[-k]z^{-k}

By change of variables z=(u^{-1})^{\ast} and k=-l for the sum of the previous equation can be written as:
P_{yx}(z)=R_{xx}(z)\sum\limits_{l=-\infty}^{\infty}h^{\ast}[l](u^{-l})^{\ast}
=R_{xx}(z)(\sum\limits_{l=-\infty}^{\infty}h[l]u^{-l})^{\ast}
=R_{xx}(z)(H(u))^{\ast}
=R_{xx}(z) \left[H(u=\frac{1}{z^{\ast}})\right]^{\ast}
=R_{xx}(z) \left[H(\frac{1}{z^{\ast}})\right]^{\ast}

The last equation proves the second part of [1, p. 54 eq. (3.51)]. By similar reasoning the third relation can also be shown to be
P_{yy}(z)=\mathcal{Z}\{r_{yy}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\star h[k]\star r_{xx}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\}\mathcal{Z}\{ h[k]\}\mathcal{Z}\{r_{xx}[k]\}
=\left[H(\frac{1}{z^{\ast}})\right]^{\ast}H(z)R_{xx}(z),

which concludes the proof. QED.

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