Solution:
Let $D$ be the mid-point of $BC$; $M$ that of $HK$; and $T$ that of $OH$. Then $PM$ is perpendicular to $HK$ and $PT$ is perpendicular to $OH$. Since $Q$ is the reflection of $P$ in $HO$, we observe that $P, T, Q$ are collinear, and $PT = TQ$. Let $QL$, $TN$ and $OS$ be the perpendiculars drawn respectively from $Q$, $T$ and $O$ onto the altitude $AK$. (See the figure.)

We have $LN = NM$, since $T$ is the mid-point of $QP$; $HN = NS$, since $T$ is the mid-point of $OH$; and $HM = MK$, as $P$ is the circumcentre of $KHO$. We obtain
$$LH + HN = LN = NM = NS + SM$$
which gives $LH = SM$. We know that $AH = 2OD$. Thus
$$\begin{aligned} AL = AH & - LH = 2OD - LH = 2SK - SM = SK + (SK - SM) = SK + MK \\ & = SK + HM = SK + HS + SM = SK + HS + LH = SK + LS = LK \end{aligned}$$
This shows that $L$ is the mid-point of $AK$ and hence lies on the line joining the mid-points of $AB$ and $AC$. We observe that the line joining the mid-points of $AB$ and $AC$ is also perpendicular to $AK$. Since $QL$ is perpendicular to $AK$, we conclude that $Q$ also lies on the line joining the mid-points of $AB$ and $AC$.

Remark: It may happen that $H$ is above $L$ as in the adjoining figure, but the result remains true here as well. We have $HN = NS$, $LN = NM$, and $HM = MK$ as earlier. Thus $HN = HL + LN$ and $NS = SM + NM$ give $HL = SM$. Now $AL = AH + HL = 2OD + SM = 2SK + SM = SK + (SK + SM) = SK + MK = SK + HM = SK + HL + LM = SK + SM + LM = LK$. The conclusion that $Q$ lies on the line joining the mid-points of $AB$ and $AC$ follows as earlier.