Let PC=r and AO=R,
Connecting PC, the chord AB of ⊙O cuts ⊙P at point C, so AB⊥PC,
If it is OD⊥AB, then od∨PC.
And ∵AB∨OP,
∴OD=PC=r,
∫ The area of the shaded part is 9π,
∴πR2-πr2=9π, that is, R2-r2=9,
So ad = $ \ sqrt {9} $ = 3.
∵OD⊥AB,
∴AB=3×2=6.
So choose C.