Ridiculous length!

We produce $AP$ to meet $BC$ at $X$ . Now $\triangle API \sim \triangle DIX$ . So,

$\dfrac{AP}{DX} = \dfrac{AI}{IX} = \dfrac{AC}{CX} = \dfrac{AB}{BX} = \dfrac{AC-AB}{CD + DX - BD + DX} = \dfrac{AC-AB}{2}$

Now, we know that $\dfrac{AC}{\sin\angle B} = \dfrac{AB}{\sin\angle C} = 2R$ , where $R$ = circumradius of $\triangle ABC$

Therefore, $AC = 2R \sin\angle B$ and $AB = 2R\sin\angle C$ . So ,

$AP = 2R(\sin\angle B - \sin\angle C) \times \frac{1}{2} = R \times (\sin\angle B - \sin\angle C)$

We need to compute $R$ . We take $O$ , the circumcentre of $\triangle ABC$ and drop $OD \perp BC$ and join $I$ and $B$ . Using $\cos\angle BOD = \cos 60^{\circ} = \dfrac{1}{2}$ and $OD = \frac{1}{2} \times AH = 5$ , we have $R = 10$

Thus $AP = R \times (\sin\angle B - \sin\angle C) = 10 \times \dfrac{3}{5} = \boxed{6}$