解：$(1)∠P=90°+ \frac {1}{2} ∠A$
理由如下：∵$DP，$$CP $分别平分$∠ADC$和$∠ACD$
∴$∠PDC=\frac {1}{2} ∠ADC，$$∠PCD= \frac {1}{2} ∠ACD$
∴$∠P=180°-∠PDC-∠PCD=180°- \frac {1}{2} (∠ADC+∠ACD)=180°- \frac {1}{2} (180°-∠A)=90°+\frac {1}{2} ∠A$
$(2)∠P= \frac {1}{2} (∠A+∠B)$
理由如下：∵$DP，$$CP $分别平分$∠ADC$和$∠BCD$
∴$∠PDC=\frac {1}{2} ∠ADC，$$∠PCD= \frac {1}{2} ∠BCD$
∴$∠P=180°-∠PDC-∠PCD=180°- \frac {1}{2} (∠ADC+∠BCD)=180°- \frac {1}{2} (360°-∠A-∠B)=\frac {1}{2} (∠A+∠B)$
