Letting $P$ be a $p$-group and $\Phi(P)$ be the Frattini subgroup of $P$ (the intersection of all maximal subgroups), the challenge is "Prove that $P/N$ is elementary abelian implies $\Phi(P)≤N$" (from Dummit and Foote 6.1.26b). The previous exercise has already established $P/\Phi(P)$ is elementary abelian since $P'≤\Phi(P)$ and $\langle~x^p~|~x∈P~\rangle≤\Phi(P)$.
The first step I took here was proving $P/N$ is elementary abelian if and only if $\langle~x^p,~P'~|~x∈P~\rangle ≤ N$, so it seemed natural to prove that $\Phi(P)=\langle~x^p,~P'~|~x∈P~\rangle$. I believe $\Phi(P)/P'=(P/P')^p$ (where $(P/P')^p$ is the image of the $p$-power map) would imply $\Phi(P)P'=\langle~x^p~|~x∈P~\rangle P'=\langle~x^p,~P'~|~x∈P~\rangle$, which would imply $\Phi(P)=\langle~x^p,~P'~|~x∈P~\rangle$ as desired, but I can't find much way to proceed further. The $\supseteq$ direction is clear, but I can't find the reverse containment. Am I going at all the right way about this?