Let O and $$\mathrm{O}^{\circ}$$ be the centres of circles P and C respectively.

The length of $$\mathrm{AB}=24$$ and as $$O O^{\prime}$$ is perpendicular to AB , we get $$\mathrm{AN}=\mathrm{NB}=12$$.

If we join AO and AO ', we get two right triangles ANO and ANO ' respectively.






In triangle ANO , using Pythagoras theorem, we get $$\mathrm{ON}=\sqrt{\mathrm{AO}^2-\mathrm{AN}^2}=\sqrt{15^2-12^2}=\sqrt{81}=9$$ (Pythagoras theorem - Hypotenuse $${ }^2=$$ Perpendicular $$^2+$$ Base $$^2$$ )

Similarly in triangle $$\mathrm{ANO}^{\prime}$$, we get $$\mathrm{O}^{\prime} \mathrm{N}=\sqrt{\mathrm{AO}^{\prime 2}-\mathrm{AN}^2}=\sqrt{13^2-12^2}=\sqrt{25}=5$$

So, the distance between the centres of two circles is $$\mathrm{ON}+\mathrm{O}^{\prime} \mathrm{N}=9+5=14$$.
Hence the answer is (D).