I know that the PGF of a shifted geometric distribution is $\frac{zp}{1-z(1-p)}$ where $z$ is the observed value of the random variable $Z$ and $0<p<1$. I am facing difficulties finding $G^{''}_Z(z)|_{z=1}$ in order to get the variance given as $G^{''}_Z(z)|_{z=1} + G^{'}_Z(z)|_{z=1} - (G^{'}_Z(z)|_{z=1})^2 = \frac{1-p}{p^2}$.
A detailed expression of $ G^{''}_Z(z)|_{z=1}$ will help.
Simply deriving the PGF and applying the formula for the variance.
Let's set $q=1-p$ to semplify the notation:
$$G'=\frac{p[1-zq]+zpq}{(1-zq)^2}=\frac{p}{(1-zq)^2}=p(1-zq)^{-2}$$
$$G''=2pq(1-zq)^{-3}$$
Now simply sustituting you get
$$G'']_{z=1}+G']_{z=1}-(G']_{z=1})^2=\frac{2pq}{p^3}+\frac{p}{p^2}-(\frac{p}{p^2})^2=\frac{2(1-p)}{p^2}+\frac{p}{p^2}-\frac{1}{p^2}=\frac{2-2p+p-1}{p^2}=\frac{1-p}{p^2}$$
as known also by other ways.
Usually to get the simple moments Moment Generating Function is used, say
$$M_{z}(t)=\frac{pe^t}{1-(1-p)e^t}$$
instead of Probability Generating Function...but the result is the same