Question: Let $ABC$ be an acute angled triangle with orthocentre $H$ and circumcircle $\Gamma$. Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let $AH$ intersect $\Gamma$ again at $P\neq A$. Let $PE$ intersect $\Gamma$ again at $Q\neq P$. Prove that $BQ$ bisects segment $EF$.
Solution: Refer to the following diagram for a better understanding of the solution.
Let us assume that $M$ is the midpoint of $EF$. Also let that $AH$ intersects $BC$ at $D$.
Now let that $\angle BCP=\gamma\implies \angle BAP=\gamma\implies \angle FEB=\gamma\implies \angle FCB=\gamma.$ Hence, $\angle BCP=\angle HCD=\gamma.$ Hence, $\Delta HCD\cong \Delta PCD.$ Thus, $HD=DP\implies HP=2HD.$
Now let that $\angle FEH=\theta\implies \angle FCB=\theta\implies \angle HED=\theta.$
Again let that $\angle FBE=\alpha\implies \angle FCE=\alpha\implies \angle HDE=\alpha.$
Thus $\Delta FBE\sim \Delta HDE.$ Hence we have $$\frac{HP}{HE}=\frac{2HD}{HE}=\frac{2FB}{FE}=\frac{2FB}{2FM}=\frac{FB}{FM}.$$ We also have $\angle PHE=\angle DHE=\angle BFE=\angle BFM$.
Hence, $\Delta FBM\sim \Delta HPE$.
Now let, $\angle ABM=\beta \implies \angle FBM=\beta\implies \angle HPE=\beta \implies \angle APQ=\beta\implies \angle ABQ=\beta.$ Hence, $\angle ABM=\angle ABQ=\beta.$
Thus, the points $B,M,Q$ are collinear and $M$ lies on $FE$, which implies that $M$ is the intersection point of $FE$ and $BQ$. And, we have $FM=ME$. Hence, we can conclude that $BQ$ bisects $FE$.
Hence, we are done.
Is this solution correct? I'm sure that this problem has quite a few alternate solutions. Can someone suggest me an alternate solution?
Yes, your solution is correct. This problem was 2019 ELMO Shortlist G1, and can be found here. The solution you found is one of the most elementary solutions, but many others can be found at that linked page. One of my favorites uses Pascal's Theorem (from projective geometry) to get a nice concurrence relation.