1. Identify the key points and properties:
- Given triangle $\triangle ABC$ with $AB = 10$ and $AC = 16$.
- $I$ is the incenter of $\triangle ABC$.
- $P$ is the intersection of the tangents to the circumcircle of $\triangle BIC$ at $B$ and $C$.
- $PA = 8$.

2. **Understand the role of point $P$:**
- $P$ is the midpoint of the arc $BC$ that contains $A$ on the circumcircle of $\triangle ABC$.
- This means $P$ lies on the external angle bisector of $\angle BAC$.

3. **Apply $\sqrt{bc}$ inversion:**
- Perform a $\sqrt{bc}$ inversion centered at $A$.
- This inversion maps $P$ to a point $P'$ on line $BC$ such that $P'A = 20$ (since $PA = 8$ and the inversion scales distances by a factor of $\sqrt{bc}$).

4. Use the angle bisector theorem:
- Since $P'$ lies on the external angle bisector of $\angle BAC$, we have:
$$\frac{P'B}{P'C} = \frac{AB}{AC} = \frac{10}{16} = \frac{5}{8}$$
- Let $P'B = 5x$ and $P'C = 8x$.

5. Apply the Power of a Point theorem:
- The Power of a Point theorem states that for a point $P'$ on the external angle bisector:
$$P'A \cdot P'A' = P'B \cdot P'C$$
- Here, $P'A = 20$ and $P'A' = 28$ (since $P'A' = PA + AA' = 8 + 20 = 28$).
- Therefore:
$$20 \cdot 28 = 5x \cdot 8x$$
- Simplifying:
$$560 = 40x^2 \implies x^2 = 14 \implies x = \sqrt{14}$$

6. **Calculate the length of $BC$:**
- Since $BC = P'B + P'C = 5x + 8x = 13x$:
$$BC = 13 \cdot \sqrt{14} = 13\sqrt{14}$$

The final answer is $\boxed{13\sqrt{14}}$