Solution 1. Let $PA,PB,PC$ cut $BC,CA,AB$ at $X,Y,Z,$ respectively. Then $\angle YCZ=\angle ZBY$ $\Longrightarrow$ $BCYZ$ is cyclic and $\angle XBY=\angle YAX$ $\Longrightarrow$ $ABXY$ is cyclic. Since $P$ is on radical axis $BY$ of $\odot(BCYZ)$ and $\odot(ABXY),$ then $CAZX$ is also cyclic. Hence $\angle BXZ=\angle BAC=\angle CXY$ and $\angle AXY=\angle ABY=\angle ACZ=\angle AXZ$ $\Longrightarrow$ $XA,BC$ bisect $\angle YXZ$ $\Longrightarrow$ $AX \perp BC$ and similarly, $BY \perp CA,$ $CZ \perp AB$ and the conclusion follows. Solution 2. If $Q$ is the isogonal conjugate of $P$ WRT $\triangle ABC,$ we have $\angle CBQ=\angle ABP=\angle ACP=\angle BCQ$ $\Longrightarrow$ $QB=QC$ and $\angle ABQ=\angle CBP=\angle CAP=\angle BAQ$ $\Longrightarrow$ $QA=QB$ $\Longrightarrow$ $QA=QB=QC$ $\Longrightarrow$ $Q$ is circumcenter of $\triangle ABC$ $\Longrightarrow$ $P$ is orthocenter of $\triangle ABC.$