A normal subgroup intersects the center of the $p$-group nontrivially

If $G$ is a finite $p$-group with a nontrivial normal subgroup $H$, then the intersection of $H$ and the center of $G$ is not trivial.

Perhaps a slightly less computationally-intensive argument is to simply note that since $H$ is normal, it must be a union of conjugacy classes. Each conjugacy class of $G$ has $p^i$ elements for some $i$; since $H$ contains at least one conjugacy class with $p^0 = 1$ elements (the class of the identity), and $|H|\equiv 0 \pmod{p}$, it must contain other classes with just one element, which must be classes of central elements of $G$.

$H$ is normal, consider $G$ acting on $H$ by conjugation.

The class equation yields $$ \left| H \right| = \left| H^G \right| + \sum_i [G:Stab_{h_i}] \, , $$ where $ H^G = \{ h \in H \mid \ ghg^{-1} = h , \ \forall g \in G \} $ are the fixed points and $ Stab_{h_i} = \{ g \in G \ \mid \ gh_ig^{-1} = h_{i} \}\leqslant G $ is the stabilizer of a $h_i \in H$. Observe that in this case $ H \cap Z(G) = H^G $.

$p$ divides $\left| H \right|$ and $[G:Stab_{h_i}]$ for every non trivial orbit, so it divides $\left| H^G \right|$. In particular, $H^G$ is not empty, so there is an element of $H$ that is also in the center of $G$.