$N$ is the midpoint of $OH$, so $AN \perp OH \iff AO = AH$. Therefore, $2R \cos A = R$ and so we get $\angle A = 60^\circ$. Now, $AN$ will also be the angle bisector of $\angle HAO$, therefore it is the angle bisector of $\angle BAC$, and hence $P$ is the midpoint of arc $BC$ not containing $A$. But this means that $P$ is the reflection of $O$ across $BC$, since if $D$ is the midpoint of $BC$, then $OD = R \cos A = R/2$.

Let $(ABC)$ be the unit circle $A=a$, $B=b$, $C=c$ and $P=p$. It´s well-know that $H=a+b+c$ and $N=\frac{a+b+c}{2}$. Since $N$ is the feet of perpendicular from $O=0$ in $AP$ which is a chord inscribed in the unit circle, we get that $N=\frac{a+b+c}{2}=\frac{1}{2}(a+p+O-ap\bar{O})=\frac{1}{2}(a+p) \Rightarrow p=b+c$ Therefore, $B+C=O+P$ so $BCOP$ is a paralelogram so $P$ is the reflection of $O$ by the line $BC$. $\blacksquare$

Rijul saini wrote: $N$ is the midpoint of $OH$, so $AN \perp OH \iff AO = AH$. Therefore, $2R \cos A = R$ and so we get $\angle A = 60^\circ$. But what if $\angle A = 120^\circ$?