There exist infinite solvable $p$-groups with trivial centre. (Use a hint.)

Recall that Robinson defines the (restricted) wreath product $H\wr K$ to be the split extension of the restricted direct product (I will borrow the notation from Hungerford) ${\prod\limits_{k\in K}}^*H$ of $|K|$ copies of $H$ by $K$. The restricted direct product is the set of (set theoretic) functions $f\colon K\to H$ of finite support, with coordinatewise multiplication. The linked to question contains the explicit description, so I won't repeat it here.

In particular, in the wreath product $H\wr K$, the subgroup ${\prod\limits_{k\in K}}^*H$ is normal, and the quotient of $H\wr K$ by this subgroup is isomorphic to $K$.

In the case at hand, $H$ and $K$ are abelian; thus, $H\wr K$ is metabelian (extension of an abelian group by an abelian group) and thus solvable. If $H\cong \mathbb{Z}_p$ and $K$ is an (infinite) elementary abelian $p$-group, it is not hard to verify that every element has order a power of $p$, so $H\wr K$ is a $p$-group.

When $K$ is infinite and $H$ is abelian, you already proved that $Z(H\wr K)=\{1\}$. Thus, $\mathbb{Z}_p\wr E$ where $E$ is an infinite elementary abelian group is (i) metabelian, hence solvable; (ii) a $p$-group; (iii) with trivial center.