You can use the result of $1$ to see that $P$ is a $p$-sylow subgroup of finite group $G$. Let $|G|=p^{\alpha}m,(p,m)=1$.
$H\leq G$ so $|H|=p^{s}n,~ s\leq\alpha,~~n\mid m$.
$P$ is a sylow-$p$ subgroup of $H$ so $|P|=p^s$.
$N_G(P)\subseteq H$ so $|N_G(P)|=p^sk,~~k\mid n$.
For all above $m,~n$ and $k$ are free of $p$ that is $(p,m)=1,~(p,n)=1,~(p,k)=1$.
Now apply the congruent relation resulted by $1$ for what we have:
$$[G:P]\equiv [N_G(P):P]~~~(\text{mod}~~ p)$$ or since $G$ is finite:
$$|G|/|P|\equiv |N_G(P)|/|P|~~~(\text{mod}~~ p)$$
or $$p^{\alpha}m/p^s\equiv p^s.k/p^s~~~(\text{mod}~~ p)$$ or
$$p^{\alpha-s}m\equiv k~~~(\text{mod}~~ p)$$
The latter makes a contradiction unless $\alpha=s$ and $m=k$. This is what you wanted in $2$.
Personally, I don't believe that the theory of central series is more elementary than Sylow theory, and in my experience students find Sylow theory easier, probably because it does not involve technical calculations with quotient groups.
But if you insist, I think you can do it this way. By using induction on $|G|$, we can assume that $Z(G)=1$. Also all maximal subgroups $M$ of $G$ are normal, so they have prime index $p$. It is not hard to show that $M$ satisfies the normalizer condition, so $M$ is nilpotent by induction. So $Z(M) \ne 1$. Let $N$ be a minimal normal subgroup of $G$ that is contained in $Z(M)$, let $G = \langle M,g \rangle$ with $g^p \in M$, and let $H = \langle N,g \rangle$.
So $Z(H) \cap N \le Z(G)$ and hence $Z(H) \cap N = 1$. Since $\langle g \rangle \cap N \le Z(H) \cap N$, we have $\langle g \rangle \cap N =1$.
Now $H$ satifies the normalizer condition, by assumption if $H=G$, and by induction otherwise. So $\langle g \rangle < N_H(\langle g \rangle)$, and hence $N_N(\langle g \rangle) \ne 1$. But $[N_N(\langle g \rangle), \langle g \rangle] \le \langle g \rangle \cap N =1$, so $N_N(\langle g \rangle) \le Z(G)$, contradiction.
Best Answer
Let $y\in N_G(H)\implies yHy^{-1}=H\implies yPy^{-1}\subset yHy^{-1}=H\implies P $ and $yPy^{-1}$ are both Sylow-$p$ subgroups of $H$,since any two Sylow $p-$ groups are conjugates hence $ P $ and $yPy^{-1}$ are conjugates in $H$.
Hence $P=x(yPy^{-1})x^{-1}$ for some $x\in H$
$\implies P=(xy)P(xy)^{-1}\implies xy\in N_G(P)\subset H$
Hence $x^{-1}(xy)=y\in H$ as a $H$ is a subgroup [$a,b\in H\implies ab^{-1}\in H$]