Solution:

We begin with the following lemma:

Lemma: Let $XYZ$ be a triangle with $\angle XYZ = 90 + \alpha$. Construct an isosceles triangle $XEZ$, externally on the side $XZ$, with base angle $\alpha$. Then $E$ is the circumcentre of $\triangle XYZ$.

Proof of the Lemma: Draw $ED \perp XZ$. Then $DE$ is the perpendicular bisector of $XZ$. We also observe that $\angle XED = \angle ZED = 90 - \alpha$. Observe that $E$ is on the perpendicular bisector of $XZ$. Construct the circumcircle of $XYZ$. Draw the perpendicular bisector of $XY$ and let it meet $DE$ in $F$. Then $F$ is the circumcentre of $\triangle XYZ$. Join $XF$. Then $\angle XFD = 90 - \alpha$. But we know that $\angle XED = 90 - \alpha$. Hence $E = F$.

Let $r_1$, $r_2$ and $r$ be the inradii of the triangles $ABD$, $CBD$ and $ABC$ respectively. Join $PD$ and $DQ$. Observe that $\angle PDQ = 90^{\circ}$. Hence

$$PQ^2 = PD^2 + DQ^2 = 2r_1^2 + 2r_2^2$$

Let $s_1 = (AB + BD + DA)/2$. Observe that $BD = ca/b$ and $AD = \sqrt{AB^2 - BD^2} = \sqrt{c^2 - \left(\frac{ca}{b}\right)^2} = c^2/b$. This gives $s_1 = cs/b$. But $r_1 = s_1 - c = (c/b)(s-b) = cr/b$. Similarly, $r_2 = ar/b$. Hence

$$PQ^2 = 2r^2 \left(\frac{c^2 + a^2}{b^2}\right) = 2r^2$$

Consider $\triangle PIQ$. Observe that $\angle PIQ = 90 + (B/2) = 135^{\circ}$. Hence $PQ$ subtends $90^{\circ}$ on the circumference of the circumcircle of $\triangle PIQ$. But we have seen that $\angle PDQ = 90^{\circ}$. Now construct a circle with $PQ$ as diameter. Let it cut $AC$ again in $K$. It follows that $\angle PKQ = 90^{\circ}$ and the points $P, D, K, Q$ are concyclic. We also notice $\angle KPQ = \angle KDQ = 45^{\circ}$ and $\angle PQK = \angle PDA = 45^{\circ}$.

Thus $PKQ$ is an isosceles right-angled triangle with $KP = KQ$. Therefore $KP^2 + KQ^2 = PQ^2 = 2r^2$ and hence $KP = KQ = r$.

Now $\angle PIQ = 90 + 45$ and $\angle PKQ = 2 \times 45^{\circ} = 90^{\circ}$ with $KP = KQ = r$.

Hence $K$ is the circumcentre of $\triangle PIQ$.

Solution:

Here we use computation to prove that the point of contact $K$ of the incircle with $AC$ is the circumcentre of $\triangle PIQ$. We show that $KP = KQ = r$.

Let $r_1$ and $r_2$ be the inradii of triangles $ABD$ and $CBD$ respectively. Draw $PL \perp AC$ and $QM \perp AC$. If $s_1$ is the semiperimeter of $\triangle ABD$, then $AL = s_1 - BD$.

But
$$s_1 = \frac{AB + BD + DA}{2}, \quad BD = \frac{ca}{b}, \quad AD = \frac{c^2}{b}$$

Hence $s_1 = cs/b$. This gives $r_1 = s_1 - c = cr/b$, $AL = s_1 - BD = c(s-a)/b$. Hence $KL = AK - AL = (s-a) - \frac{c(s-a)}{b} = \frac{(b-c)(s-a)}{b}$.

We observe that $2r^2 = \frac{(c+a-b)^2}{2} = \frac{c^2 + a^2 + b^2 - 2bc - 2ab + 2ca}{2} = (b^2 - ba - bc + ac) = (b-c)(b-a)$.

$$\begin{aligned} (s-a)(b-c) &= (s-b + b-a)(b-c) \\ &= r(b-c) + (b-a)(b-c) \\ &= r(b-c) + 2r^2 = r(b-c + c + a - b) = ra \end{aligned}$$

Thus $KL = ra/b$. Finally,
$$KP^2 = KL^2 + LP^2 = \frac{r^2 a^2}{b^2} + \frac{r^2 + c^2}{b^2} = r^2$$

Thus $KP = r$. Similarly, $KQ = r$. This gives $KP = KI = KQ = r$ and therefore $K$ is the circumcentre of $\triangle PIQ$.