normally embedded subgroups of solvable $T$-group

Definition: A subgroup $H$ of a group $G$ is said to be normally embedded in $G$ if each Sylow $p$-subgroup of $H$ is a Sylow $p$-subgroup of some normal subgroup $N$ of $G$.

Definition: A group $G$ is a $T$-group if has a transitive relation among its subgroups i.e. all subnormal subgroups of $G$ are normal in $G$. The following is a classical result given by Gaschutz on the structure of the finite solvable $T$-groups

Theorem: Let $G$ be a finite solvable $T$-group. Then

(a) $G$ contains an abelian normal subgroup $L$.

(b) every subgroup of $G/L$ is normal in $G/L$

The following is adapted from an exercise in Finite Soluble Groups by Doerk and Klaus.

A finite group $G$ is a solvable $T$-group $\iff$ if every subgroup is normally embedded in $G$.

I have managed to show the sufficiency condition. For the necessary condition, fix a subgroup $H$ of $G$. I need to show that $H$ is normally embedded in $G$. Let $P$ be a Sylow $p$-subgroup of $H$ for some prime $p$. By the theorem, we know $G$ has a normal abelian subgroup $L$. Suppose that $p$ does not divide $|L|$. Then it is clear that $P$ is a Sylow $p$-subgroup of $PL$. Moreover, $PL/L$ is normal in $G/L$, and so $PL$ is normal in $G$. In this case, $H$ is normally embedded in $G$.

I'm not sure how to proceed for the case if $p$ divides $|L|$

Let $L = R \times Q$, where $R$ is a $p'$-group and $Q$ is a $p$-group. Now $PQ$ is a $p$-group, so $P$ is subnormal in $PQ$, and hence $PR$ is subnormal in $PL$, But $PL \unlhd G$, so $PR$ is subnormal and hence normal in $G$, with $P \in {\rm Syl}_p(PR)$.