Let $BC = a$, $AC = b$, $AB = c$ and $\angle CAB = \alpha$, $\angle ABC = \beta$, $\angle BCA = \gamma$. Let $D$ be the intersection point of $BL$ and $CK$. Note that $\triangle PAQ$ is isosceles.
$$\angle BKL = \angle APK - \angle ABK = \frac{\pi - \alpha}{2} - \frac{\beta}{2} = \frac{\alpha + \beta + \gamma - \alpha - \gamma}{2} = \frac{\gamma}{2} = \frac{\angle ACB}{2}.$$
Since $\angle IKL = \angle BKL = \angle ABC/2 = \angle ACI$, the points $I, K, Q, C$ are concyclic, and hence, $\angle IKC = \angle IQC = \pi/2$. Similarly, $\angle ILB = \pi/2$. Therefore, the points $B, L, K, C$ are concyclic, and the points $I, L, D, K$ are also. Particularly, $BC$ is a diameter of the circumscribed circle of $\triangle CLK$ and $ID$ is a diameter of the circumscribed circle of $\triangle ILK$.

Using the sines law in $\triangle ILK$ we have
$$ID = \frac{IK}{\sin \angle LDK} = \frac{LK}{\cos \angle LCK},$$
and in $\triangle CLK$ we have $a = \frac{LK}{\sin \angle LCK}$.
Therefore, $ID = a \tan \angle LCK$.
Since $\sin \angle LCK = \sin \angle IQK = \sin(\alpha/2)$, we also can write that $ID = \tan(\alpha/2)$. On the other hand, $r = AQ \tan(\alpha/2)$ and $AQ = (b+c-a)/2$, where $r$ is a radius of incircle of $\triangle ABC$.

Circumcircle of $\triangle ILK$ touches the incircle of $\triangle ABC \iff$ diameter of the circumcircle of $\triangle ILK$ is equal to the radius of incircle of $\triangle ABC$.