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.