Let $G$ is a finite group. Suppose $H \unlhd G$ such that $G/H$ is supersoluble. Suppose $H \cap M$ is either $H$ or a maximal subgroup of $H$ for any maximal subgroup of $G$. Suppose that $P_{H}(M) = \langle h \in H \mid \langle h \rangle M = M \langle h \rangle \rangle = H$ for any maximal subgroup $M$ of $G$. Let $p \in \pi(G)$ be an odd prime, then $G$ is $p$-supersoluble.
For proof: Since $P_{H}(M) = H$, then $H$ is soluble, then $G$ is soluble. Now assume $G$ is not $p$-supersoluble and $G$ is a counterexample of smallest order. Let $N$ be a minimal normal subgroup of $G$, then $N$ is an elementary abelian $q$-group for prime $q$. I show $G/N$ is $p$-supersoluble group and $N$ is unique minimal normal subgroup. Now $\Phi(G) = 1$ or $\Phi(G) \neq 1$?