Index of Hall subgroup divides normalizer of center of Sylow-p subgroup

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I came across this problem recently and was unable to solve it after multiple tries. Let $ A \triangleleft G$ be a normal Hall subgroup(i.e $(ord(N),ord([G:A]))=1$) and let $P$ be a Sylow-p subgroup of $ A$. Let $m$ be the index of $A$. Show that $m \vert$ $|N(Z(P))$ where $N(Z(P))$ is the normalizer of the center of $P$.

Obviously, $P$ is also a Sylow-p subgroup of $G$ by the coprimality condition. Also, $Z(P)$ is not empty, which can be proven using the class equation.But I am unable to prove anything beyond that. Any help is appreciated.