Normalizer of Sylow $2$-subgroup of $D_{2n}$.

Question

In Normalizer of a Sylow 2-subgroups of dihedral groups it is proved that the number of Sylow $2$-subgroups of $D_{2n}$ if $2n=2^ak$ where $k$ is odd is $k$ without proving $N_G(P)=P$ if $P$ is Sylow $2$-subgroup of $D_{2n}$. I am trying to prove the fact $N_G(P)=P$ then deduce the number of Sylow$2$-subgroups is $k$. Assume$a,k>1$. If $N_G(P)$ is larger than $P$, then if $g\in N_G(P)$, the order of $g$ must divide $k$, so $g$ is in $\langle r^{2^a} \rangle$. But from here I cannot proceed. Any suggestion?