因为,∠BPA = 180°-∠BAP-∠ABP = 70° = ∠BAP ,
所以,BA = BP ,可得：△ABP是等腰三角形；
过点A作AD⊥BC于D,交CP延长线于O,连接OB；
过点B作BE⊥CP于E,则点E在CO延长线上；
AD是等腰△ABC底边上的高,可得：AD是BC的垂直平分线,
而且O在AD上,可得：OB = OC ,
∠OBC = ∠OCB = 30° ,
∠CBE = 90°-∠OCB = 60° ,
∠OBE = ∠CBE-∠OBC = 30° ；
因为,在△OBD和△OBE中,∠ODB = 90° = ∠OEB ,∠OBD = 30° = ∠OBE ,OB = OB ,
所以,△OBD ≌ △OBE ,
可得：OD = OE ,BD = BE ；
因为,在Rt△ABD和Rt△PBE中,AB = PB ,BD = BE ,
所以,△ABD ≌ △PBE ,
可得：AD = PE ；
因为,在△BOA和△BOP中,OA = AD-OD = PE-OE = OP ,BA = BP ,OB = OB ,
所以,△BOA ≌ △BOP ,
可得：∠OBA = ∠OBP = ½∠ABP = 20° ,
所以,∠PBC = ∠OBC-∠OBP = 10° .