Solution:

We get the following diagrams by taking the cross-section of the plane of the triangle.

Note that $AC = 9$ by the Pythagorean theorem. We also get that $\triangle ADP \sim \triangle ACB.$

We can see that $PCBD$ is a kite, so we know that $$DB = BC = 12,$$ which makes $AD = 15 - 12 = 3.$ Using the similar triangles, above, we get that $$\dfrac{r}{3} = \dfrac{12}{9}$$$$r = 4.$$

Let $d$ be the distance from the sphere to this plane. $d$ is also the distance from $O$ to $P.$

Once again using the Pythagorean theorem, we get that $$d = \sqrt{6^2 - 4^2} = 2\sqrt{5}.$$

Thus, D is the correct answer.