Nice Problem! Please refrain from giving spoilers about problems in the thread name though. We immediately observe that $L$ is the Miquel Point of quadrilateral $APQB$. Now, we have the following claim. Claim : The circles centered at $A$ with radius $AP$ and $B$ with radius $BQ$ intersect on $AB$. We prove this as follows. Let $R$ be a point on $AB$ and $P'$ the intersection of $AC$ and the circle centered at $A$ and radius $AR$. Now, let $Q' = (P'IC) \cap \overline{BC}$. It suffices to show that $BR=RQ'$. For this, note that since $AR=AP'$ and $\measuredangle IAR = \measuredangle P'AR$ we have that $\triangle IAR \cong \triangle P'AI$. Thus, $IR=IP'$. Further, \[\measuredangle Q'P'I = \measuredangle Q'CI = \measuredangle ICP' = \measuredangle IQ'P'\]from which it follows that $IP'=IQ'$. Thus, $I$ must be the center of $\triangle RP'Q'$. Now, we can note that \[2\measuredangle Q'RB = 2\measuredangle Q'RA = 2\measuredangle Q'RP' + 2\measuredangle P'RA = \measuredangle Q'IP' + \measuredangle P'AR = \measuredangle Q'CP' + \measuredangle P'AR' = \measuredangle BCA + \measuredangle CAB = \measuredangle CBA\]Thus, it is clear that $BR=BQ'$ as desired. Thus, as $\omega$ varies, for all $P$ and $Q$ intersections of $\omega$ with $AC$ and $BC$ we have that the circles centered at $A$ with radius $AP$ and $B$ with radius $BQ$ intersect on $AB$. From the above claim, it follows that $AP+BQ=AB$. Now, armed with this information, we can attack the problem. Since $L$ is the Miquel point of $APIQ$, it follows that $\triangle LAP \sim \triangle LBQ$. Note that then, \[\frac{AK}{KB} = \frac{AL}{LB}=\frac{AP}{BQ}\]But, since $AP+BQ=AB=AK+KB$ it follows that $AK=AP$ and $KB=BQ$. Thus, we have that \[2\measuredangle PKQ = 2\measuredangle PKA + 2\measuredangle BKQ = \measuredangle PAK + \measuredangle KBQ = \measuredangle CAB + \measuredangle ABC = \measuredangle ACB\]Thus, $\measuredangle PKQ$ clearly does not depend on $P$ and $Q$ and indeed as $\omega$ varies, $\angle PKQ$ is constant.

Here's an elementary approach. Clearly $IP=IQ$ from the bisector $CI$ in $\omega$. Our aim is to show that $IK=IP$, as it would follow that $I$ is the circumcenter of triangle $PKQ$, whence $\angle PKQ=\frac{1}{2} \angle PIQ =90^{\circ}-\frac{1}{2}{\angle ACB}$. Let the angle bisectors $CI$ and $LK$ intersect at the midpoint $T$ of the arc $\widehat{AB}$ from $k$, not containing $C$. We have $\angle TAK=\angle TAB=\angle BCT=\angle ACT=\angle ALT$, thus $\triangle AKT\sim\triangle LAT$ and $TA^2=TK\cdot TL$. Since $TA=TI$ from the trillium lemma, we deduce $TI^2=TK\cdot TL$ and $\triangle IKT\sim\triangle LIT$. Hence $$ IK=IT\cdot \frac{LI}{LT}=AT\cdot\frac{LI}{LT}. $$On the other hand, the circumcircles $k$ and $\omega$ yield $\angle LPI=180^\circ-\angle LCI=\angle LAT$ and $\angle PLI=\angle PCI=\angle ALT$. Hence $\triangle LAT\sim\triangle LPI$ and so $$ \frac{LI}{LT}=\frac{PI}{AT} $$which completes the proof. P.S. @above, what spoilers have I given in the title, they are just words from the problem statement, no hints??