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Note first that $ABCD$ is concyclic since it is an isosceles trapezoid. We define $X$ to be the intersection of $\odot (DQT)$ and $\odot (ABCD)$, and it now suffices to prove that $BRQX$ and $BPSX$ are cyclic quadrilaterals. We have that $$\begin{array}{rl}\angle BXQ& =\angle BXD+\angle DXQ=(\pi -\angle BCD)+\angle DTQ\\ & =\angle RCD+\angle CDR=\angle RCT+\angle CTR=\pi -\angle CRT\end{array}$$ and hence $BRQX$ is a cyclic quadrilateral. Now observe that $$\begin{array}{rl}\angle AXQ& =\angle AXD+\angle DXQ=(\pi -\angle ACD)+\angle DTQ\\ & =\angle PCD+\angle CTP=\angle PCT+\angle CTP\\ & =\pi -\angle CPT=\pi -\angle APQ\end{array}$$ so $APQX$ is concyclic, and it now follows that: $$\angle SBX=\angle DBX=\angle DAX=\angle QAX=\angle QPX=\angle SPX$$ Hence, $BPSX$ is a cyclic quadrilateral.