This an excellent problem! Let $H$ be the orthocenter of triangle $ABC$ and $\ell$ be the line through $H$ parallel to $BC$. Let $D, E$ be the touch-points of the incircle and $C$-excircle (henceforth $\omega_C$) on side $AB$, respectively. Let $E'$ be antipode of $E$ in $\omega_C$. Let $X$ be the exsimilicenter of $(CH)$ and $\omega_C$. Note that $C, D, E'$ are collinear by dilation at $C$. The negative dilation $\mathcal{H}: \omega_C \mapsto (CH)$ has pivot $X$; and $\mathcal{H}: BC \mapsto \ell$, $E \mapsto H$ and $E' \mapsto C$ since $BC$ and $\ell$ are tangents to $\omega_C$ and $(CH)$ respectively, and $EE', CH$ are diameters of these circles, respectively. Let the lines $XP$ and $XQ$ meet line $\ell$ at $Y, Z$ respectively, and line $CX$ meet $\ell$ at $H'$. Note that $(CH)$ is the excircle opposite vertex $X$ in triangle $XYZ$, so $H$ is the extouch point, while $H'$ is the intouch point on $YZ$. Therefore, the $X$-medians of triangle $XYZ$ and $XHH'$ coincide. Apply $\mathcal{H}$ to the last result; we get that the $X$-medians of triangle $XDE$ and $XPQ$ coincide, as desired. $\blacksquare$

Stats- 1 student solved this during the contest.

Let $T$ be the center of $\odot(CKL)$ and let $I_c$ be the center of $\omega.$ Let the incircle of $\triangle ABC$ and $\omega$ touch $AB$ at $D$ and $E$ respectively. Let $E'$ be the antipode of $E$ on $\omega.$ Let the common tangents to $\odot(CKL)$ and $\omega$ meet at $J$—the insimilicenter of the two circles. Since the orthocenter of $\triangle ABC$ is the antipode of $C$ on $\odot(CKL)$, it follows that $CT \perp AB.$ Therefore $CT \parallel I_cE'.$ The homothety mapping $\overline{CT} \mapsto \overline{I_cE'}$ is clearly negative, so it must be centered at the insimilicenter $J.$ Therefore $J \in CE'.$ It is known from homothety that $C, D, E'$ are collinear, so in fact $J \in CD.$ By the Dual of Desargues' Involution Theorem for quadrilateral $CAEB$ and point $J$, there is an involution swapping $(JD, JE); (JA, JB); (JP, JQ).$ Thus, there is an involution on line $AB$ swapping $(D, E); (A, B); (P, Q).$ Since $D, E$ are symmetric in $\overline{AB}$, this involution must be isotomic conjugation w.r.t. $\overline{AB}.$ Therefore $P, Q$ are isotomic w.r.t. $\overline{AB}$ as well.

I had the same solution as Kapil; see here for a very similar problem.

Perhaps the same solution as #4? anantmudgal09 wrote: Let $AK$ and $BL$ be the altitudes of an acute-angled triangle $ABC$, and let $\omega$ be the excircle of $ABC$ touching side $AB$. The common internal tangents to circles $CKL$ and $\omega$ meet $AB$ at points $P$ and $Q$. Prove that $AP =BQ$. Proposed by I.Frolov Let $\omega$ touch $\overline{AB}$ at point $S$; let $Z$ be the insimilicenter of $\omega$ and $\odot(CKL)$. By homothety, $\overline{CZ}$ passes through the antipode of $S$ in $\omega$; hence $T=\overline{CZ} \cap \overline{AB}$ is symmetric to $S$ in midpoint of $\overline{AB}$. Now consider the inconic $\omega$ for the quadrilateral $CASB$ and point $Z$; by Dual of Desrague Involution Theorem, $(P,Q), (A,B), (T,S)$ are pairs of an involution. However $(T,S), (A,B)$ determine only one involution: reflection in midpoint of $\overline{AB}$; hence $AP=BQ$. $\blacksquare$

Amazing The statement is true for any circle with its centre on the $C$-altitude, passing through $C$. I'll prove the more general result. Let $D$ be a point on the $C$-altitude of $\triangle ABC$, let $\Gamma$ be the circle centred $D$ with radius $CD$, and let $\omega$ be the $C$-excircle of $\triangle ABC$. The common internal tangents of $\omega$ and $\Gamma$ intersect $\overline{AB}$ at point $P$ and $Q$. Prove that $AP = BQ$. (Assuming that their internal common tangent exists) Proof. Let $Z$ be the insimilicenter of $\omega$ and $\Gamma$, $I_c$ be $C$-excentre of $\triangle ABC$, $X$ be the $C$-extouch point, $X'$ be the antipode of $X$ in $\omega$, $Y$ be the $C$-intouch point of $\triangle ABC$ and $Y'$ be the $Z$-intouch point of $\triangle ZPQ$. Note that $CD\parallel X'I_c$. Thus, $CZX'$ are collinear. Now note that $Y$ lies on $CX'$ since $C$ is exsimilicenter of the incircle of $\triangle ABC$ and $\omega$. Also, note that $\omega$ is the $Z$-excircle. Thus, by similar reasoning as above, $Y'$ lies on $ZX'$. However, we have $ZX' \equiv CX'$, thus, $Y' \equiv Y$. Now since $X$ and $Y$ are the $C$ - intouch and extouch points of $\triangle ZPQ$, respectively, they are isotomic in $\overline{PQ}$. However, $X$ and $Y$ are isotomic in $\overline{AB}$. Thus, we must have $AP = BQ$.