= Solution
Conditioning on $N$ and using the <moment-generating function> of a unit exponential variable,
$$
\begin{aligned}
\mathbb E(e^{\theta Y})
&=\sum_{n=1}^\infty p(1-p)^{n-1}
\left(\frac1{1-\theta}\right)^n\\
&=\frac{p}{1-\theta}
\sum_{m=0}^\infty
\left(\frac{1-p}{1-\theta}\right)^m
=\boxed{\frac{p}{p-\theta}},
\end{aligned}
$$
where the geometric <series> converges precisely when $\theta<p$. This is the moment-generating <function> of an exponential variable of rate $p$, so the <geometric sum of exponential variables> satisfies
$$
\boxed{Y\sim\operatorname{Exp}(p)}.
$$
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