$E \equiv AC \cap BD.$ From cyclic $EPBQ, ABCD, EQCR$ $\Longrightarrow$ $\angle EQP = \angle EBP = \angle ABD = \angle ACD = \angle ECR = \angle EQR$ $\Longrightarrow$ $QE$ bisects $\angle PQR.$ Similarly, $RE, SE, PE$ bisect $\angle QRS, \angle RSP, \angle SPQ$ $\Longrightarrow$ $PQRS$ is tangential with an incircle $(E).$