∵∠ACD是△ABC的外角∴∠ACD=∠A+∠ABC=70°+∠ABC∵CP是∠ACD的平分线∴∠DCP= 1 2 ∠ACD= 1 2 （70°+∠ABC）=35°+ 1 2 ∠ABC∵BP是∠ABC的平分线∴∠CBP= 1 2 ∠ABC∵∠DCP是△BCP的外角∴∠DCP=∠CBP+∠P35°+ 1 2 ∠ABC= 1 2 ∠ABC+∠P∴∠P=35°．故答案为：35°．
