In △COD, ∠ C+∠ D+∠ COD = 180,
∠∠AOB =∠COD，
∴∠a+∠b=∠c+∠d；
(2) As shown in Figure 2, ∫AP and CP share ∠BAD and ∠BCD equally,
∴∠ 1=∠2,∠3=∠4,
∠∠2+∠B =∠3+∠P，
∠ 1+∠P=∠4+∠D，
∴2∠P=∠B+∠D，
∴∠p= 12(∠b+∠d)= 12×(36+ 16)= 26；
① As shown in Figure 3, ∫AP bisects ∠BAD's external angle ∠FAD, and CP bisects ∠BCD's external angle ∠BCE,
∴∠ 1=∠2,∠3=∠4,
∴∠PAD= 180 -∠2，∠PCD= 180 -∠3，
∫∠P+( 180-∠ 1)=∠D+( 180-∠3)，
∠P+∠ 1=∠B+∠4，
∴2∠P=∠B+∠D，
∴∠p= 12(∠b+∠d)= 12×(36+ 16)= 26；
② As shown in Figure 4, ∫AP bisects ∠BAD's external angle ∠FAD, and CP bisects ∠BCD's external angle ∠BCE,
∴∠ 1=∠2,∠3=∠4,
∴( 180-2∠ 1)+∠b =( 180-2∠4)+∠d，
In quadrilateral APCB, (180-∠1)+∠ p+∠ 4+∠ b = 360,
In quadrilateral APCD ∠ 2+∠ P+( 180-∠ 3)+∠ D = 360,
∴2∠P+∠B+∠D=360，
∴∠p= 180- 12(∠b+∠d)；
③ As shown in Figure 5, ∫AP bisects ∠BAD, and CP bisects ∠BCD's outer corner ∠BCE,
∴∠ 1=∠2,∠3=∠4,
∫(≈ 1+∠2)+∠B =( 180-2∠3)+∠D，
∠2+∠P=( 180 -∠3)+∠D，
∴2∠P= 180 +∠D+∠B，
∴∠P=90 + 12(∠B+∠D)。