 Solution

For a complex irreducible representation, every central element acts as a scalar by Schur lemma. If the representation is faithful, these scalars embed $Z$ in ${\mathbb{C}}^{\ast }$. Every finite subgroup of ${\mathbb{C}}^{\ast }$ is cyclic: if its order is $m$, all its elements satisfy $z^m=1$, and the $m$th roots of unity form a cyclic group. Thus $Z$ is cyclic.

The trivial group satisfies the conclusion immediately. Otherwise let $G$ be a finite $p$-group with cyclic centre, and decompose a faithful representation as a direct sum of irreducibles using Maschke's theorem. The nontrivial cyclic $p$-group $Z$ has a unique subgroup $C$ of order $p$. Some component does not kill $C$, since otherwise the whole representation would kill it. If that component had a nontrivial kernel $N$, the kernel would be a normal subgroup and the allowed intersection theorem gives $N\cap Z\ne 1$. Every nontrivial subgroup of cyclic $Z$ contains $C$, contradicting the choice of component. That irreducible component is faithful.

Finally the regular representation is faithful, since left multiplication by a nonidentity element does not fix the basis vector at the identity. A cyclic centre therefore supplies a faithful irreducible component of it. The first argument proves the converse, giving a finite $p$-group has a faithful irreducible complex representation exactly when its centre is cyclic.