Let $Q$ be the midpoint of $BC$, $A'$ be the reflection of $A$ into $O$, where $O$ is the circumcenter of triangle $ABC$, and $H$ be the orthocenter of $ABC$.

Quadrilateral $BHCA'$ is a parallelogram, therefore $A'H$ passes through $Q$. It also intersects the circumcircle again at $R$ where
$$\widehat{HRA} = \widehat{A'RA} = 90^\circ.$$
Which means $R$ lies on circle with diameter $AH$. Consider three circles, circle with diameter $AH$, circumcircle of triangle $ABC$ and circle passing through $B, H, C$. The radical axes of these three circles are lines $AR, EF, BC$, and they are concurrent at a point $K$.
$AH, HQ$ are altitudes of triangle $AKQ$, therefore $H$ is also the orthocenter of triangle $AKQ$. So if $S$ is the intersection of $KH, AQ$, we have $\widehat{ASH} = 90^\circ$ which means $S$ also lies on the circle with diameter $AH$. Now we have
$$KH \cdot KS = KR \cdot KA = KB \cdot KC.$$
Therefore $B, H, S, C$ lie on a circle. Note that $\widehat{SEF} = \widehat{SHC} = \widehat{SBC}$ and similarly, $\widehat{SFE} = \widehat{SCB}$. So we conclude that triangles $SEF$ and $SBC$ are similar. Also, since $AN$ is the angle bisector of $\widehat{A}$, we have
$$\frac{ME}{MF} = \frac{AE}{AF} = \frac{AB}{AC} = \frac{BN}{CN}$$
$$\Rightarrow \widehat{\Delta SME} \sim \widehat{\Delta SNB} \Rightarrow \widehat{SME} = \widehat{SNB} = \widehat{SNK}.$$
Therefore points $K, M, S, N$ lie on a circle. Let $P'$ be the second intersection point of this circle with $AQ$. It's easy to see that $\widehat{P'MK} = \widehat{P'SK} = 90^\circ$ and similarly $\widehat{P'NK} = 90^\circ$. Hence $P \equiv P'$ and the claim of the problem is concluded. ■