$(1)$证明：∵$∠A+∠B+∠AOB = 180°，$$∠C+∠D+∠COD = 180°$
∴$∠A+∠B+∠AOB=∠C+∠D+∠COD$
∵$∠AOB=∠COD，$∴$∠A+∠B=∠C+∠D$
解：$(2)$∵$AP，$$CP $分别平分$∠BAD，$$∠BCD$
∴$∠BAP=∠P AD，$$∠BCP=∠P CD$
$ $由$(1)$的结论，得$∠P+∠BCP=∠ABC+∠BAP ①，$
$∠P+∠P AD=∠ADC+∠P CD ②$
①+②，得$2∠P+∠BCP+∠P AD=∠BAP+∠P CD+∠ABC+∠ADC$
∴$2∠P=∠ABC+∠ADC$
∵$∠ABC = 36°，$$∠ADC = 16°，$∴$2∠P=36°+16°=52°$
∴$∠P = 26°$
$ (3) ∠P = 90°+\frac 12(∠B+∠D)，$理由如下：
∵直线$AP $平分$∠BAD，$$CP $平分$∠BCD$的外角$∠BCE$
∴$∠P AB=∠P AD，$$∠P CB=∠P CE$
∴$2∠P AB+∠B = 180°-2∠P CB+∠D$
∴$180°-2(∠P AB+∠P CB)+∠D=∠B$
∵$∠P+∠P AD=∠P CB+∠AOC=∠P CB+∠B + 2∠P AD$
∴$∠P=∠P AD+∠B+∠P CB=∠P AB+∠B+∠P CB$
∴$∠P AB+∠P CB=∠P-∠B$
∴$180°-2(∠P-∠B)+∠D=∠B，$即$∠P = 90°+\frac 12(∠B+∠D)$
