Nice problem, Let $CD\cap AB=S$,$KP\cap AB=T$, Note that $2SK=LP$,$LP\bot AB$,so $TS=SL$, We have $LC\bot CK$,so $SLCK$ is cyclic Which means that $\angle TKS = \angle SKL =\angle SCL=\angle SCB- \dfrac{1}{2}\angle C$ Since $\measuredangle (DK,AB)=90^{\circ}- \dfrac{1}{2}\angle C+\angle SCB$ So $\measuredangle (DK,AB) + \angle TKS=180^{\circ}\Rightarrow D,T,K$ is collinear Finally $\angle DPL = \angle TKS = \angle SKL = \angle DCL$ We get $DLCP$ is cyclic