Let $V$ be the circumcenter of triangle $KPH$ and $O$ the circumcenter of triangle $ABC$ It is a well-known lemma that $P$ is the reflection of $H$ with respect to $BC$ hence $BC$ is the side bissector of $KP$ hence $V$ lies on $BC$ Since $PQ$ is the radical axes of $(ABC)$ and $(KPH)$ we have $PQ$ is perpendicular to $OV$ thus $TN$ is parallel to $OV$ Let $KO \cap TN=T'$ Then we have $$\frac{KO}{OT'}=\frac{KV}{VN}=1 \quad(KN \quad is \quad the \quad diameter)$$ Hence $$KO=OT'$$ We know that $$BD=CK \implies BK=CD$$ And since $BO=CO$ and $\angle OBK=\angle OBC= \angle OCB= \angle OCD$ We get $\triangle OBK \cong \triangle OCD$ (By SAS) Hence $OK=OD$ Thus $OK=OD=OT'$ so $O$ is the circumcenter of $(KDT')$ And since $O$ lies in $KT'$ we get $\angle T'DK=90$ hence $T=T'$ So $O$ lies on $TK$

Let $(KPH) \cap NT = S$. Clearly, $\angle NSK = 90$, so $PQ // SK$ and $SDKT$ is cyclic. The perpendicular bisector of $PQ$ passes through $O$, so the perpendicular bisector of $SK$ also passes through $O$, therefore $SO = OK$. The perpendicular bisector of $BC$ passes through $O$, so the perpendicular bisector of $DK$ also passes through $O$, therefore $DO = OK$. This shows us that $O$ is the center of $(SDTK)$, and this implies $\overline{T, O, K}$.