It's clear that $PB = PA$. Let $\angle PBA = \angle PAB = \alpha$. Since $ABPQ$ is inscribed in a circle $\angle AQP = 180^{\circ} - \alpha$. Let $R = SP\cap CD$. Then $$\angle QSP = \angle ESR = 360^{\circ} - (\angle SED + \angle EDR + \angle DRS) = 360^{\circ} - 108^{\circ} - 108^{\circ} - 90^{\circ} = 54^{\circ} $$$$\angle QPS = 180^{\circ} - (\angle SQP + \angle QSP) = 180^{\circ} - 180^{\circ} + \alpha - 54^{\circ} = \alpha - 54^{\circ} $$$$\angle ASP = 180^{\circ} - \angle QSP = 180^{\circ} - 54^{\circ} = 126^{\circ} $$$$\angle SAP = \angle EAB - \angle PAB = 108^{\circ} - \alpha$$$$\angle APS = 180^{\circ} - (\angle ASP + \angle SAP) = 180^{\circ} - 126^{\circ} - 108^{\circ} + \alpha = \alpha - 54^{\circ} $$So $\angle QPS = \angle APS$ as required.