Expected value of a Delta function times a function
If $h(t)$ is continuous at $a$, then $h(t)\delta(t-a) = h(a)\delta(t-a)$. Since the 
$\operatorname{sinc}$ function is
continuous everywhere, 
$$\begin{align}
Y(t) &amp;= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(t)\\
&amp;= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(n\tau)\\
&amp;= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)\operatorname{sinc}(n)\\
&amp;= Z_0\delta(t)
\end{align}$$
since $\operatorname{sinc}(n) = \begin{cases}1,&amp;n = 0,\\0,&amp; n\neq 0.\end{cases}$
You (or your textbook's authors) probably meant to write something else.
You sure you are not missing a $\star$ indicating a convolution between
that $\delta(t-n\tau)$ and $h(t)$?