First off, no exist another point $X$ on $AE$ such that $\angle ABX=\angle ACX$, it this because the isogonal conjugate of $X$ lies on perpendicular bisector of $BC$, so the locus of $X$ is a circumconic of $\triangle ABC$, so no exist three points ($A,E,X$) collinear on this conic. Let $H$ be the orthocenter of $\triangle$, and $C_1 \equiv HC \cap AB, B_1 \equiv BH \cap AC$ Let $O_1$ be the symmetric of the circumcenter of $\triangle ABC$ on $BC$, So $AO_1$ is the Euler Line of $\triangle BHC \Rightarrow E \in AO_1$ $F = \odot (BO_1C) \cap AO_1, (F \not= O_1) \Rightarrow \angle BFO_1=\angle O_1FC=90^{\circ}-\alpha$ If $\odot (AFB)$ cut again $AC$ at $B_2 \Rightarrow \angle BB_2B_1=90^{\circ}-\alpha$ Analogously $\odot (AFC)$ cut again $AB$ at $C_2 \Rightarrow \angle CC_2C_1=90^{\circ}-\alpha \Rightarrow BB_2C_2C$ is cyclical, By radical axis on $\odot (ABF), \odot (ACF)$ and $\odot (BB_2C_2C)$ we get $K \equiv AE \cap BB_2 \cap CC_2$ is $K$ lies on segment $AE$ then $K \equiv P$ as desired.

See Easy Geometry or angles PBA and PCA are equal

I came up with an amazing solution Solution: Let $Q$ be isogonal conjugate point of $P$ wrt $\triangle ABC$, since $\angle ABP=\angle ACP=x$ then we have $\angle QBC=\angle QCB=x$ or $Q$ is on perpendicular bisector of $BC$ (1) On the other hand, $AP,AQ$ are isogonal conjugate rays emitted from $A$, we arrive at $AQ$ is passing through the center of $(OBC)$, in another words, $Q$ is the center of $(OBC)$ (see (1)) Of course, it 's apparent to see that, with the implement of angle-chasing, we can easily arrive at $\angle QBC=\angle QCB=90-2\angle BAC$ or $x=90-2\alpha$, as desired!