∵AB∥CD∴∠BEF+∠DFE=180°又∵∠BEF的平分线与∠DFE的平分线相交于点P∴∠PEF= 1 2 ∠BEF，∠PFE= 1 2 ∠DFE∴∠PEF+∠PFE= 1 2 （∠BEF+∠DFE）=90°∵∠PEF+∠PFE+∠P=180°∴∠P=90°．故答案为90°．
