$\DeclareMathOperator\Syl{Syl}$Let $G$ be a finite group, $\Phi(G)$ is the Frattini subgroup of $G$. And $G/\Phi(G)$ is a simple group. Let $P\in \Syl_{p}(\Phi(G))$, where $p\in \pi(\Phi(G))$ and  $\pi(\Phi(G))$ is the set of prime divisors  of $|\Phi(G)|$.

Suppose that every subgroup $H$ of $P$ of order $p$ is normal in $G$.

Let $P_1\in \Syl_{p}(G)$. Then for any subgroup $H$ of $P$ with $|H|=p$, $H\leq Z(P_1)$. So $P_1\leq C_G(H)\unlhd G$.

Under the above conditions, can we use the simplicity of $G/\Phi(G)$ to conclude that $H\leq Z(G)$?