Let $(OKM)$ intersect $BC$ again at $T'$ clearly $OKT'=90^{\circ}$. See that $\angle KMO=\angle KAH=\angle KT'O$, also $\angle OKT'=\angle HKA=90^{\circ}$, from here we notice the spiral similarity that sends $T'O \mapsto AH$ centered at $K$ and it's known that $D=AT'\cap OH$ is the intersection of $(OKT'M), (AEKHF)$, and then $AT'\perp OH$. Let $G=AT'\cap EF$, and $I$ the midpoint of $AC$, not hard to see that $AO\perp EF$, and clearly $OI\perp AC$, with this we have $GDOAO\cap EF$ and $EIOAO\cap EF$ are both cyclic and hence $GDIE$ is cyclic too. Now $\angle GIE=\angle GDE=\angle AFE=\angle ACB$ and $GI\| BC$, and $G$ is midpoint of $AT'$. Now a bit of angle chasing leads to $AQ\| EF$ and $P$ midpoint of $QT' \implies T\equiv T'$

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