The group $G$ is a nontrivial <finite p-group>, so the <nontrivial center of a finite p-group> gives $Z(G)\ne\{e\}$. If $Z(G)\ne G$, then $Z(G)$ is already a nontrivial proper <normal subgroup>, so $G$ is not a <simple group>. If $Z(G)=G$, then $G$ is <abelian group>[abelian]. By <Cauchy's theorem for finite groups>, it has a subgroup of order $p$; every subgroup of an abelian group is normal, and this subgroup is nontrivial and proper because $|G|=p^4$. Thus in every case
$$
\boxed{G\text{ is not simple}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 1 / 10G / a / Solution

Every nontrivial <finite p-group> has a nontrivial <center of a group>[center]. The conjugation action partitions the group into conjugacy classes whose noncentral sizes are positive powers of $p$; the class equation then implies $|Z(G)|\equiv|G|\equiv0\pmod p$.

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 Incoming links: Nontrivial center of a finite p-group