Power Spectral Density of a Filter
A digital filter doesn't have power spectral density, a signal has. I guess what you want is the transfer function of this filter, i.e., $H(z)=Y(z)/X(z)$.
Let's add two intermediate signal $u(n)$ and $v(n)$ in the diagram.

We have:
$$
y(n) = v(n-1) + du(n) \tag{1}
$$
$$
u(n)=v(n-1)+x(n)\tag{2}
$$
$$
v(n) = x(n) -du(n) \tag{3}
$$
Take Z-transform and we get
$$
Y(z) = V(z)z^{-1} + dU(z) \tag{4}
$$
$$
U(z) = V(z)z^{-1}+X(z)\tag{5}
$$
$$
V(z)=X(z)-dU(z)\tag{6}
$$
Furthermore, we can derive that
$$
V(z)=X(z)-d\Big[V(z)z^{-1} + X(z)\Big]
$$
$$
\frac{V(z)}{X(z)} = \frac{1-d}{1+dz^{-1}}
$$
$$
Y(z)=V(z)z^{-1} + d\Big[V(z)z^{-1}+X(z) \Big] =(1+d)V(z)z^{-1} + dX(z)
$$
Thus,
$$
H(z) = \frac{Y(z)}{X(z)} = (1+d) \frac{V(z)}{X(z)} z^{-1} + d = (1+d) \frac{(1-d)z^{-1}}{1+dz^{-1}} +d
$$