Given an arbitrary triangle $ABC$, with $\angle A = 60^o$ and $AC < AB$. A circle with diameter $BC$, intersects $AB$ and $AC$ at $F$ and $E$, respectively. Lines $BE$ and $CF$ intersect at $D$. Let $\Gamma$ be the circumcircle of $BCD$, where the center of $\Gamma$ is $O$. Circle $\Gamma$ intersects the line $AB$ and the extension of $AC$ at $M$ and $N$, respectively. $MN$ intersects $BC$ at $P$. Prove that points $A$, $P$, $O$ lie on the same line.