Let $p$ be a prime and $G$ a group. A subgroup $M\le G$ is a $p$-local subgroup of $G$ if there exists a nontrivial $p$-subgroup $P$ of $G$ such that $N_G(P)=M$.
If for each prime divisor $p$ of $|G|$, all the $p$-local subgroups are subnormal in $G$, can we say that all the subgroups of $G$ are subnormal?
Yes, this is true because if $P$ is a Sylow $p$-subgroup of $G$ then $N_G(P)$ is self-normalising in $G$ by an application of Frattini's argument, i.e. it is its own normaliser. Since $N_G(P)$ is subnormal by assumption, it follows that $N_G(P) = G$ and thus $P$ is normal in $G$. Since this holds for all primes $p$ we have that $G$ is nilpotent, hence every subgroup of $G$ is subnormal.
Observe that it was not necessary to assume that all $p$-local subgroups are subnormal. The same conclusion can be reached by merely assuming that the normalisers of the Sylow subgroups of $G$ are subnormal in $G$.