A minimal normal subgroup of a $p$-solvable group is either a $p$-group or $p'$-group.
Let $G$ be a finite group and $p$ be a prime. Then $G$ is $p$-solvable if
(i) $G$ has a subnormal series where each factor group is either a $p$-group or $p'$-group.
(ii) Every composition factors of $G$ is either a $p$-group or $p'$-group.
$A$ is a minimal normal subgroup of $G$ if $A\lhd G$ and for any normal subgroup $B$ of $G$ which is contained in $A$, either $1=B$ or $B=A$.
Since the condition (1) is not true for every subnormal series, the subnormal series $1\lhd A\lhd G$ cannot be used to lead to conclusion.
We know that there is a composition series containing $A$, say $$1\lhd\dots\lhd H\lhd A\lhd K \lhd \dots \lhd G$$
By assumption, we know that $A/H$ and $K/A$ are either a $p$-group or $p'$-group.
Here I can't proceed further because $H$ is not necessarily a normal subgroup of $G$, so the property of $A$ being a minimal normal subgroup cannot be applied to $H$.
Best Answer
It is easy to see that in $p$-solvable groups $G$ either $O_p(G)$ or $O_{p'}(G)$ is non-trivial (or both). If $N$ is a minimal normal subgroup of a $p$-solvable group $G$, then $N$ itself is $p$-solvable, hence $O_p(N) \gt 1$ or $O_{p'}(N) \gt 1$. But these subgroups are characteristic in $N$ and, since $N$ is normal, they are normal in $G$. So $N=O_p(N)$ or $N=O_{p'}(N)$ and you are done.