Summations in Z-Transform
This may help: the result is the cascade of an FIR filter with coefficients $h(n)$ with the reverse conjugate filter with coefficients $h^*(N-n)$. If we can limit $H(z)$ to be the unit circle (for computing the frequency response), then
$$H(z^*)|_{z=e^{j\omega}} = H(z^{-1})|_{z=e^{j\omega}}$$
since
for $z = e^{j\omega}$, $z^* = e^{-j\omega}$
And therefore
$$H^*(1/z)|_{z=e^{j\omega}}= H^*(z^*)|_{z=e^{j\omega}}$$
And
$$H^*(1/z^*)|_{z=e^{j\omega}}= H^*(z)|_{z=e^{j\omega}}$$
So
$$Y(z)|_{z=e^{j\omega}} = H(z)H^*(z)|_{z=e^{j\omega}} = |H(z)|^2|_{z=e^{j\omega}}$$

Also from the link the OP provided to me in the comments below, it is clear that the problem is dealing with the two sided Z-transform and the full problem has a solution given as
$$H(z)H^*(1/z^*)\Phi_{xx}(z) = \frac{\sigma_v^2}{(1-az^{-1})(1-a^*z)}$$
And we see from the earlier solution in the link that
$$H(z) = \frac{1}{1-az^{-1}}$$
Therefore
$$H^*(z) = \frac{1}{1-a^*(1/z^*)}$$
And
$$H^*(1/z) = \frac{1}{1-a^*(z^*)}$$
And
$$H^*(1/z^*) = \frac{1}{1-a^*z}$$
And
$$H(z)H^*(1/z^*)=\frac{1}{(1-az^{-1})(1-a^*z)}$$
Note confirming my earlier result, when we restrict z to be the unit circle (for determining the magnitude response):
$$H^*(z)|_{z=e^{j\omega}} = \frac{1}{1-a^*(1/z^*)}|_{z=e^{j\omega}}= \frac{1}{1-a^*(1/e^{-j\omega})} = \frac{1}{1-a^*(e^{j\omega})} = \frac{1}{1-a^*z}|_{z=e^{j\omega}}$$
Also to avoid any confusion, earlier in the linked solutions where it says &quot;from 7... we arrive at (2.55) as follows&quot; it gives a result that indicates
$$\Phi_{xx}(z) = \Phi^*_{xx}(1/z^*)$$
But this is not generally true for any z transform, (we cannot conclude that $H(z) = H^*(1/z^*)$ for all z) but it is in this case specific to the autocorrelation function $\phi_{xx}(k)$ since that function is Hermitian:
$$\phi^*_{xx}(k) = \phi_{xx}(-k)$$