求解下面这道题

解：过点D作DM⊥BC延长线，垂足M∵AB⊥BC∴∠B=∠DMP=90°∵PA⊥PD∴∠APD=90°∴∠A+∠APB=∠APB+∠DPM=90°∴∠A=∠DPM=21°∵PA=PD∴△APB≌△PDM∴PB=DM，AB=PM=BC∴PB+PC=PC+CM∴PB=CM=DM∴∠DCM=45°=∠D+∠DPM∴∠D=24°