Let $M$ be the circumcentre of $\triangle PKH$. Note that the reflection of $H$ in $BC$ lies on the circumcircle $\omega$. (One way to see this is to note that $\angle BHC = \angle BHD + \angle DHC = \angle BCA + \angle ABC$.) Therefore this reflection is the point $P$. So the reflection in $BC$ transforms $\triangle PKH$ into itself, so $M$ lies on $BC$ and therefore is the midpoint of $KN$.

Let $O$ be the centre of $\omega$. Note that $P$ and $Q$ lie on both $\omega$ and the circumcircle of $\triangle PKH$, so $OM$ is the perpendicular bisector of $PQ$. Therefore $OM \perp PQ \perp TN$, so $OM \parallel TN$. As $M$ is the midpoint of $KN$, the line $OM$ is the midsegment of $\triangle KNT$ parallel to $TN$. Thus $OM$ passes through the midpoint of $KT$, say $O'$.

Let $m$ be the perpendicular bisector of $KD$. It is a midsegment in $\triangle KDT$, because it passes through the midpoint of $KD$ and is parallel to $DT$. Therefore $m$ also passes through $O'$. As $|BD| = |KC|$, $m$ is also the perpendicular bisector of $BC$. So $m$ also passes through $O$, as $BC$ is a chord of $\omega$. Since $OM$ and $m$ both pass through both $O$ and $O'$, we conclude that $O' = O$ if $OM$ and $m$ do not coincide. Therefore $KT$ passes through $O$ if $OM$ and $m$ do not coincide.

Finally, to complete the proof, we show that $OM$ and $m$ indeed do not coincide. Let $\ell$ be the line through $N$ perpendicular to $PQ$. Note that $H$ lies in the interior of $\triangle ABC$, as it is an acute-angled triangle. It follows that $D \neq N$ and therefore that $\ell$ does not coincide with $AD$. Since $D \neq N$ and $M$ is the midpoint of $KN$, $M$ is not the midpoint of $KD$. So $M$ is not the midpoint of $BC$. Therefore $OM$ is not perpendicular to $BC$ and therefore not parallel to $m$; in particular, $OM$ and $m$ do not coincide. $\square$