Let $H_a$ be the $A-$altitude of $ABC$, $\Omega$ be the circumcircle of $PNH_a$,and $M$ be the midpoint of $EF$. Since $EF$ is the antiparallel of $BC$, we must have the $A-$symmedian of $ABC$ is the $A-$median of $AEF$, thus $\overline{A-M-L}$ is collinear, because $NM$ is the perpendicular bisector of $EF$, $M$ lies on $\Omega$. Let $AM$ intersect $\Omega$ again at $L'$, using Power of Point $A$ with respect to $\Omega$ and the nine-point circle of $ABC$, we get $$ AM. AL' = AN . AH_a = AH_b. \frac{AC}{2} $$Reflect $A$ with respect to $L'$ and let the reflection be $A'$, we have $$AH_b . AC = AM (2AL') = AM . AA' $$Thus $H_bCA'M$ is cyclic, we get $$ \measuredangle AA'C = \measuredangle MA'C = \measuredangle MH_bC = \measuredangle H_cH_bC = \measuredangle H_cBC = \measuredangle ABC $$Hence, $A'$ lies on $(O)$, it follows that $OL'$ is the perpendicular bisector of $AA'$ and $L' = L$, and thus we get $ \angle NLP = \angle NH_aP = \angle AH_aC=90^{\circ}$.