Let the $Q$-antipode wrt $\Gamma$ be $R$; then $RLTP$ is cyclic by PoP at $D$, so $\angle{DTL} = \angle{KPQ}$, so it suffices to show that $KTPQ$ is cyclic, but this is just true since $\angle{TKD} = \angle{TDA} = \angle{TQR}$. $\square$

Let $S$ be the middle of the arc $AB$ of the circle $\Gamma$, different from $Q$. According to Archimedes' lemma, the straight $TD$ passes through the point $S$. In addition, by the degree of the point (or by virtue of symmetry with respect to $AB$), we obtain that the points $P, T, L, S$ lie on the same circle. Then $\angle DTL = \angle SPL$. And then we want to check that the points $P, Q, K, T$ lie on the same circle. Is this equivalent to the equality of angles $\angle TQS =^? \angle TKD = \angle TDB$. But this is already obvious from the equality of arcs $AS, BS$.

Let ${V}$ be the second intersection of circle $\Gamma$ with $PQ$. From homothety of center $T$ that turns circle $KTD$ into circle $\Gamma$ we get that $T,D,V$ are collinear. Let $QV\cap AB={N}$.Now we have $\angle QTV=\angle DNV=90^{\circ} $ and $\angle NVD=\angle TVQ$. This implies that $\angle TQV=\angle NDV=180^{\circ}-\angle TDB=\angle TKL$ .By PoP we have $DL\cdot PD= DB\cdot AD=TD\cdot DV$ so $PTLV$ is cyclic. This implies that $\angle QVT=\angle KLT$. Now we have that triangles $QTV$ and $KTL$ are similar. So we have that $\angle QTV=\angle KTL$ and this implies that $\angle QTK=\angle DTL$.So the problem is proved.

My solution from the contest: Let's first denote with $Z$ the midpoint of the arc $AB$ not containing $Q$. From the Death Star lemma we have that $D, T, Z$ lie on the same line. Since $\delta$ and $\Gamma$ are touching $\Rightarrow$ From homothety with center $T$ we send $K$ to $F$ with $F \in \Gamma \Rightarrow FZ \parallel KD \parallel PD$ $\Rightarrow \angle FZP = \angle ZPD$. We also know that $\angle FZP = \angle FZQ = \angle FTQ = \angle KTQ$ from cyclic $ ZTQF$. Now let's notice that from power of a point $D$ we have $ ZD.DT  = AD.BD$ (from $\Gamma$) $= PD.DL $ (from $\Omega$) $\Rightarrow TPZL$ is cyclic $\Rightarrow \angle ZPD =  \angle ZPL =  \angle ZTL = \angle DTL$. Finally, we obtain $\angle DTL = \angle ZPD = \angle FZP = \angle QTK$. $\square$

Let $R\neq Q$, $S$ be the intersection of $PQ$ and $\Gamma$, $AB$ respectively. Since by tangency there is a homothety on $T$ mapping $\delta$ to $\Omega$, $R$ passes through $TD$, and by radical axis $RLTP$ is cyclic. Since $\angle DTQ=\angle RTQ=90^\circ=\angle DSQ$, $TDSQ$ is cyclic, and $\angle TKP=\angle TKD=\angle TDA=\angle TQS=\angle TQP$, so $TKQP$ is cyclic. Now we have $\angle LTD=\angle LTR=\angle LPR=\angle KPQ=\angle KTQ$.

Let $R$ be the antipode of $Q$. By the Shooting Lemma, $T$, $D$, and $R$ are collinear. Notice that $TD\cdot TR=AD\cdot DB=PD\cdot DL$ so $PTLR$ is concyclic. The rest is an easy angle chase $$\angle QTK=90^{\circ}-\angle KTD=90^{\circ}-\angle PDB=\angle DPR=\angle DTL$$