If $N\lhd G$ then $n_p(G/N) \leq n_p(G)$.

Even something stronger is true.

Proposition Let $G$ be a finite group, $N \unlhd G$, then $n_p(G/N) \mid n_p(G)$.

One proof (it can also be shown in a different way, see here) hinges on the following lemma of which the easy proof can be found for example in the book of I.M. Isaacs, Finite Group Theory, remarks made on page $23$. Note that if $H \leq G$ and $N \unlhd G$, then $N \subseteq HN \subseteq N_G(HN)$. Also, since $N$ is normal, $N_G(H) \subseteq N_G(HN)$.

Lemma Let $G$ be a group, $H \leq G$, $N \unlhd G$, then $$N_{G/N}(HN/N)=N_G(HN)/N.$$ Now, every Sylow $p$-subgroup of $G/N$ is of the form $PN/N$, for some $P \in Syl_p(G)$. Hence, $n_p(G/N)=|G/N:N_{G/N}(PN/N)|= \text{(by the lemma) }= |G/N:N_G(PN)/N|=|G:N_G(PN)|$. But, $N_G(P) \subseteq N_G(PN)$, hence $n_p(G/N)=|G:N_G(PN)| \mid |G:N_G(PN)| \cdot |N_G(PN): N_G(P)|= |G:N_G(P)|=n_p(G)$ and we are done.

The key idea is the correspondence isomorphism theorem: there is a bijection between subgroups of $G/N$ and subgroups of $G$ containing $N$.

Let $f:G\rightarrow H$ be a surjective homomorphism with kernel $N$. If $P$ is a Sylow subgroup of $G$, then $f(P)$ is a Sylow subgroup of $H$ ($[G:PN]=[H:f(P)]$, and $f(P)$ is a $p$-group). If $Q$ is a Sylow subgroup of $H$, there is a subgroup $N\le K\le G$ such that $f(K)=Q$. If $L$ is a Sylow subgroup of $K$, then also $f(L)=Q$ (by the first part, and since $f(L)\le f(K)=Q$).

That means that every Sylow subgroup of $H$ has a Sylow preimage in $G$, so $n_p(H)\le n_p(G)$.