Let $M$ the midpoint of $PQ$, since $ST\parallel PQ$ we get $B$, $K$ and $M$ are collinear. Since $Q=AD\cap BC$ and $P=AB\cap CD$ it's well-know $P$ and $Q$ are conjugated respect to $\odot (ABCD)$ $\Longrightarrow$ the circumference of diameter $PQ$ is orthogonal to $\odot (ABCD)$ , let $X$ be a point in $\odot (ABCD)$ such that $MX$ is tangent to $\odot (ABCD)$ $\Longrightarrow$ $MQ^2=MP^2=MX^2=MK.MB$ $\Longrightarrow$ $\measuredangle MQK=\measuredangle MBQ$ and $\measuredangle MPK$ $=$ $\measuredangle MBP$ $\Longrightarrow$ $\measuredangle TKS+\measuredangle SBT=180^{\circ}$ hence $BSKT$ is cyclic

Suppose $L$ is midpoint of $PQ$. Since $ST\parallel PQ$ and $K = PT\cap QS$, then $BK$ passes through $L$. $\boxed{\textbf{Claim.}\; \odot (L, LP) \;\text{and}\; \odot (O) \; \text{are orthogonal}}$

Sketch

Hence ${\cal P}_{L, (O)} = \overline{LK}.\overline{LB} = LQ^{2}$. Since $\angle BLQ = \angle QLK$ then we have $\triangle BQL \sim \triangle QKL$. Thus, $\angle TSK = \angle KQL \;(\text{Since} \; ST \parallel PQ)\; = \angle QBL = \angle TBK$ and we're done.