(a) Let $A'$ be the reflection of $A$ over $BC$, and let $M$ be the second intersection point of line $AI$ with the circumcircle $\Gamma$ of $ABC$. Since $\triangle ABA'$ and $\triangle AOC$ are both isosceles triangles, and $\angle ABA' = 2\angle ABC = \angle AOC$, they are similar to each other. Similarly, $\triangle ABI_A$ and $\triangle AIC$ are also similar. Thus we have
$$\frac{AA'}{AI_A} = \frac{AA'}{AB} \cdot \frac{AB}{AI_A} = \frac{AC}{AO} \cdot \frac{AI}{AC} = \frac{AI}{AO}.$$
Combined with $\angle A'AI_A = \angle IAO$, this shows that $\triangle AA'I_A$ and $\triangle AIO$ are similar.

Let $P'$ be the intersection point of line $AP$ with $OI$. Using directed angles (denoted $\angle^*$), we compute:
$$\begin{aligned} \angle^* MAP' &= \angle^* I'_A AI_A = \angle^* I'_A AA' - \angle^* I_A AA' \\ &= \angle^* AA' I_A - \angle^* (AM, OM) \\ &= \angle^* AIO - \angle^* AMO = \angle^* MOP'. \end{aligned}$$
So $M, O, A, P'$ are concyclic.

Let $R, r$ denote the circumradius and inradius of $\triangle ABC$ respectively. Then we have
$$IP' = \frac{IA \cdot IM}{IO} = \frac{IO^2 - R^2}{IO},$$
which is clearly independent of $A$. Therefore, the intersection point of $BP$ with $OI$ is also $P'$, so $P = P'$.
This proves that point $P$ lies on line $OI$.

(b) By Poncelet's Porism, the other tangent lines drawn from $X,Y$ to the incircle of $ABC$ meet at a point $Z$ on $\Gamma$. Let $T$ be the point of tangency of $XY$ with the incircle, and let $D$ be the midpoint of segment $XY$. We compute
$$\begin{aligned} OD &= IT \cdot \frac{OP}{IP} = r\left(1 + \frac{OI}{IP}\right) \\ &= r\left(1 + \frac{OI^2}{OI \cdot IP}\right) \\ &= r\left(1 + \frac{R^2 - 2Rr}{R^2 - IO^2}\right) \\ &= r\left(1 + \frac{R^2 - 2Rr}{2Rr}\right) = \frac{R}{2} = \frac{OX}{2} \end{aligned}$$
From this we know $\angle XZY = 60^\circ$, so $\angle XIY = 120^\circ$.
The proof is complete.