We have $\angle{PSD} = \angle{PBD} = \angle{PAC} = \angle{PAS}$. So, $PS$ is tangent to the circumcircle of $\triangle{ADS}$ so $PS^2 = PD \cdot PA = PC \cdot PB$ so $PS$ is tangent to $\triangle{BCS}$ then $\angle{PBS} = \angle{PSC} = \angle{PTS}$ so $PS = PT$. Analogously, we deduce that $PV = PU$. Let the diagonals $AC$ and $BD$ intersect at point $Q$. Notice that because $QA \cdot QC = QB \cdot QD$ the power of $Q$ w.r.t circumcircle of $\triangle{PAC}$ and $\triangle{PBD}$ are equal, so $QU.QV = QS.QT$ so $STUV$ is cyclic. Now, it is clear that due to equality $PS = PT$ and $PV = PU$ we have that $P$ is the intersection of the midperpendicular of $UV$ and $ST$ but because $STUV$ is cyclic, that means $P$ is the circumcenter of $STUV$. Thus, $PS = PT = PU = PV$.