Let $G$ be a finite group such that all subgroups of prime-power order are pronormal in $G$. If $M$ is a normal $p$-subgroup of $G$ then all prime-power order subgroups of $G/M$ are pronormal in $G/M$.
I need to show that $G$ is supersolvable. The author ams.org/journals/proc/1969-020-01 says that it follows easily from the above but I'm not too familiar with the characterisations of finite supersolvable groups to get a conclusion from the hypothseis.
You should perhaps have mentioned that the author has already shown at this stage that $G$ is solvable.
Let $N$ be a minimal normal subgroup of $G$. Then, since $G$ is solvable, $N$ is an elementary abelian $p$-group for some prime $p$. Let $M$ be a subgroup of $N$ of order $p$.
Then $M$ is pronormal in $G$ and, so for any $g \in G$, $M$ and $M^g$ are conjugate in $\langle M,M^g \rangle$. But $\langle M,M^g \rangle \le N$, which is abelian, so $M=M^g$ and $M \unlhd G$ (so in fact $N=M$).
Since all prime-power order subgroups of $G/M$ are pronormal in $G/M$, $G/M$ is supersolvable by induction. Hence $G$ is supersolvable.