Let $I_B, J_B, I_C, J_C$ be the incenters of $\triangle ABK$, $\triangle ABL$, $\triangle ACK$, $\triangle ACL$ respectively.

We begin with the following claim which does not depend on the existence of tangency points $P$ and $Q$.
Claim — Lines $BC$, $I_BJ_C$, $J_BI_CI_C$ meet at a point $R$ (possibly at infinity).
Proof. Note that
$$(BI; I_BJ_B) = \frac{\sin \angle I_BAB}{\sin \angle I_BAI} \div \frac{\sin \angle J_BAB}{\sin \angle J_BAI} = \frac{\sin \frac{1}{2}\angle BAK}{\sin \frac{1}{2}\angle CAK} \div \frac{\sin \frac{1}{2}\angle BAL}{\sin \frac{1}{2}\angle CAL}.$$
Similarly
$$(CI; J_CI_CI_C) = \frac{\sin \angle J_CAC}{\sin \angle J_CAI} \div \frac{\sin \angle I_CAC}{\sin \angle I_CAI} = \frac{\sin \frac{1}{2}\angle CAL}{\sin \frac{1}{2}\angle BAL} \div \frac{\sin \frac{1}{2}\angle CAK}{\sin \frac{1}{2}\angle BAK}.$$
Thus the cross ratios are equal. Therefore, the concurrence follows from the so-called prism lemma on $\overline{IBI_BJ_B}$ and $\overline{ICJ_CI_CI_C}$. $\square$

Claim — Line $PQ$ also passes through $R$.
Proof. Note $(BP; I_BJ_B) = -1 = (CQ; J_CI_CI_C)$, so the conclusion again follows by prism lemma. $\square$

As above, the lines $BC$, $I_B J_C$, $J_B I_C$ meet at some point $R$ (possibly at infinity). Let $\omega_1, \omega_2, \omega_3, \omega_4$ be the incircles of $\triangle ABK$, $\triangle ACL$, $\triangle ABL$, and $\triangle ACK$.
Claim — There exists an inversion $\iota$ at $R$ swapping $\{\omega_1, \omega_2\}$ and $\{\omega_3, \omega_4\}$.
Proof. Consider the inversion at $R$ swapping $\omega_1$ and $\omega_2$. Since $\omega_1$ and $\omega_3$ are tangent, the image of $\omega_3$ is tangent to $\omega_2$ and is also tangent to $BC$. The circle $\omega_4$ is on the correct side of $\omega_3$ to be this image. $\square$
Claim — Circles $\omega_1, \omega_2, \omega_3, \omega_4$ share a common radical center.
Proof. Let $\Omega$ be the circle with center $R$ fixed under $\iota$, and let $k$ be the circle through $P$ centered at the radical center of $\Omega$, $\omega_1, \omega_3$.
Then $k$ is actually orthogonal to $\Omega$, $\omega_1, \omega_3$, so $k$ is fixed under $\iota$ and $k$ is also orthogonal to $\omega_2$ and $\omega_4$. Thus the center of $k$ is the desired radical center. $\square$
The desired statement immediately follows. Indeed, letting $S$ be the radical center, it follows that $\overline{SP}$ and $\overline{SQ}$ are the common internal tangents to $\{\omega_1, \omega_3\}$ and $\{\omega_2, \omega_4\}$.
Since $S$ is the radical center, $SP = SQ$. In light of $\angle SPI = \angle SQI = 90^\circ$, it follows that $IP = IQ$, as desired.