Let $(O)$ be the circumcircle of triangle $ABC$. $AK$ is the diameter of $(O)$. If $D$, $S$, $T$ are the reflection points of $H$ through $BC$, $CA$, $AB$, then $D$, $S$, $T$ are on $(O)$. It's easily seen that $HCKB$ is a parallelogram so $M$ lies on $HK$. Let $KH$ intersect $(O)$ again at $G$. Lines $HP$ intersects the lines $BC$, $DK$, $ST$, $GA$ at $R$, $X$, $Y$, $Z$ respectively.

Note that $EF$ is the median line of triangle $HST$ so $HY = 2HP$. Because $OH \perp YR$, according to butterfly theorem for segments $YR$ and $ZX$, it is obtained for quadrilateral $BCTS$ that $HR = HY = 2HP$. By butterfly theorem for quadrilateral $AGDK$ it is also obtained that $HZ = HX = 2HR = 2HY = 4HP$. Note that $PQ \perp HM$ so $PQ \parallel AG$. Noticing the parallel lines, it is deduced that $HA = 4HQ$, this means $QA = 3QH$. Hence the proof is done. ■