 Solution

Initially $G_1=G$ is powerfully embedded in itself. If $G_i$ is powerfully embedded in $G$, then $[G_i,G]\subseteq G_i^p$ and consequently $G_{i+1}=G_i^p$. The preliminary result says that $G_i^p$ is again powerfully embedded in $G$, completing the induction.

For a finite p-group $A$, the Frattini subgroup is $\Phi (A)=A^p[A,A]$. Here $[G_i,G_i]\subseteq [G_i,G]\subseteq G_i^p$, soSince the Frattini subgroup of a nontrivial finite p-group is proper, this series eventually reaches $1$. Thus $P_j=G^{p^j}$, with iterated power subgroups, terminates.