Singular $p$-subgroups of a finite group

Not sure if this is true in general, but it is when $\color{blue}{N \subseteq \bigcap_{P \in Syl_p(G)}N_G(P)}$, the intersection of all the normalizers of the Sylow $p$-subgroups of $G$. Call this intersection $D$. Note that $D \unlhd G$ and that it has a unique Sylow $p$-subgroup, namely $O_p(G)$, which is $\bigcap_{P \in Syl_p(G)}P$. So a "good" example for $N$ would be $N=O_p(G)$.

Anyhow, let $QN/N \in Syl_p(G/N)$, with $SN/N \subseteq QN/N$. Then $SN \subseteq QN$. But $N$ normalizes $Q$, so $Q \unlhd QN$, hence $Q$ is the unique Sylow $p$-subgroup of $QN$. But of course the $p$-subgroup $S \subseteq SN \subseteq QN$. Since by Sylow theory $S$ must be contained in some Sylow $p$-subgroup of $QN$, we get $S \subseteq Q$. And the choice of $S$ now gives $Q=P$, whence $PN/N=QN/N$.

Note (One day later ...) The general case is true, and I can only blame myself not having read the comments of @j.p.! We need the following.

Lemma Let $G$ be a a group, $H \leq G$ and $N \trianglelefteq G$. Then for any $g \in G$ we have $(H \cap N)^g=H^g \cap N$.

Proof Since $H \cap N \subseteq H$, $(H \cap N)^g \subseteq H^g$, and likewise $(H \cap N)^g \subseteq N^g=N$ (the latter equality follows from the fact that $N$ is normal. Hence $(H \cap N)^g \subseteq H^g \cap N$. Conversely, let $x \in H^g \cap N$, then $x \in N$ and $x=g^{-1}hg$ for some $h \in H$. Hence $h= gxg^{-1} \in N$, since $x \in N$ and $N$ is normal. So, $x=g^{-1}hg \in (H \cap N)^g$ and we are done.

Now assume that $SN/N \subseteq QN/N$, for some $Q \in Syl_p(G)$. Then $S \subseteq QN$. Now obviously, $Q \in Syl_p(QN)$ and hence the cardinality of Sylow subgroups in $QN$ is the same as in $G$! By Sylow and the fact that $S$ is a $p$-group, $S \subseteq Q^x$ for some $x \in QN$. But this $Q^x$ must actually be a Sylow subgroup of $G$ by the previous remarks. So, $Q^x=P$, and hence, $P \subseteq QN$. Finally we need to show that $PN=QN$ and that is where the Lemma comes in: $|QN|=\frac{|Q| \cdot |N|}{|Q \cap N|}=\frac{|Q^x| \cdot |N|}{|(Q \cap N)^x|}=\frac{|P| \cdot |N|}{|P \cap N|}=|PN|$. It follows that $PN=QN$, whence $PN/N=QN/N$ and we are done.

Additional remark It can be proved in general for a $p$-subgroup $S$, that $\#\{P \in Syl_p(G): S \subseteq P\} \equiv 1$ mod $p$. (Sketch of the proof: let $S$ act by conjugation on the set $\{P \in Syl_p(G): S \subseteq P\}$ and observe that $N_S(P)=P \cap S$.)