According to the original problem, the first circle is supposed to be tangent to $AL$ and $BL$, not $CL$. Can you please fix this in your post?

Here is a solution with Muriatic. We claim the more general statement holds: Generalization wrote: In triangle $AKL$, lines $AB$ and $AC$ are isogonal. A circle $\omega_K$ is tangent to rays $KA$ and $KL$ as well as internally tangent to $(ABC)$ at $N$, and circle $\omega_L$ is tangent to rays $LA$ and $LK$ as well as internally tangent to $(ABC)$ at $M$. The lines $KN$ and $LM$ intersect at $P$. Then, $A, P$, and the $A$-mixtilinear intouch point in $\Delta AKL$ are collinear. It is clear that, in the original problem, $AB$ and $AC$ are isogonal in $\angle AKL$, and that, since $\omega$ is the $A$-excircle of $\Delta AKL$ and thus $E$ is the $A$-extouch point, $AE$ is isogonal with the line through $A$ and the $A$-mixtilinear intouch point (for example, by $\sqrt{KL}$-inversion). So, this generalization clearly implies the original problem. Let $\gamma$ be the incircle of $\Delta AKL$. We note that $K$ is the exsimilicenter of $\gamma$ and $\omega_K$, and $N$ is the exsimilicenter of $\omega_K$ and $\Omega$, so by Monge's theorem, the exsimilicenter of $\gamma$ and $\Omega$ lies on line $KN$. Similarly, this exsimilicenter also lies on line $LM$, and it is thus $P$. Now, $A$ is the exsimilicenter of $(AKL)$ and $\Omega$ (as $AB$ and $AC$ are isogonal, $(ABC)$ and $(AKL)$ are tangent at $A$, for example by $\sqrt{KL}$-inversion), which gives by another application of Monge's theorem that $A$, $P$, and the exsimilicenter of $\gamma$ and $(AKL)$ is on line $AP$. Let $Q$ be the $A$-mixtilinear intouch point in $\Delta AKL$. As $A$ is the exsimilicenter of the $A$-mixtilinear incircle and $\gamma$, and $Q$ is the exsimilicenter of the $A$-mixtilinear incircle and $(AKL)$, we have by a fourth application of Monge's that the exsimilicenter of $\gamma$ and $(AKL)$ is on line $AQ$, so $A$, $P$, and $Q$ are collinear, finishing the proof.

Here is an alternate finish to the original problem (i.e. not the generalization) after we get that $P$ is the exsimilicenter of $(ABC)$ and the incircle $\gamma$ of $AKL$. Let $X$ be such that $XB, XC$ are tangent to $(XBC)$. We claim that $\omega$ is the incircle of $XBC$. Indeed, since $O$ is the midpoint of arc $BC$ in $(XBC)$ and $D$ lies on $XO$ so that $OD = OB = OC$, by Fact 5 $D$ is the incenter of $XBC$, so since $\omega$ is tangent to $BC$ it is tangent to all $3$ sides. Thus $X$ is the exsimilicenter of $\omega$ and $(ABC)$, so by Monge on $\gamma$, $\omega$, and $(ABC)$ we get that $A, P, X$ are collinear, so $P$ lies on the $A$-symmedian of $ABC$.

Here is a solution with Desargues' Theorem and Monge's Theorem. In the following, we use $e (\alpha, \beta)$ to denote the exsimilicenter of the circles $\alpha, \beta.$ Firstly, note that lines $AK, AL$ are reflections over line $AD$. Hence, since $AE$ is the median in $\triangle ABC$, we just need to show that $AP$ is the $A-$symmedian of $\triangle ABC.$ Let $X = BB \cap CC$ be the intersection of the tangents to $\Omega$ at $B$ and $C$. It's well-known that $AX$ is the $A-$symmedian of $\triangle ABC$, so we'll just show that $A, P, X$ are collinear. In other words, we desire to show that $LM, KN, AX$ are concurrent. Since $\angle DBC = \frac12 \angle A = \frac12 \angle XBC,$ and analogously $\angle DCB = \frac12 \angle XCB$, we have that $\omega$ is the incircle of $\triangle BCX$. In particular, $BX, CX$ are tangent not only to $\Omega$ but also to $\omega$. From here, it follows that $X = e(\Omega, \omega).$ By Desargues' Theorem on $\triangle AKL, \triangle XNM$, it suffices to show that $XM \cap AL, XN \cap AK, MN \cap KL$ are collinear. At first, these points may seem arbitrary, but actually we are only four Monge's'es away from solving the problem! By Monge's Theorem on $\gamma_1, \Omega, \omega$, we know that $e(\gamma_1, \omega)$ lies on $XM$. Hence, since it clearly lies on $AL$ (common tangent of $\gamma_1, \omega$), we have that $XM \cap AL = e(\gamma_1, \omega).$ By Monge's Theorem on $\gamma_2, \Omega, \omega$, we know that $e(\gamma_2, \omega)$ lies on $XN$. Hence, since it clearly lies on $AK$ (common tangent of $\gamma_2, \omega$), we have that $XN \cap AK = e(\gamma_2, \omega).$ By Monge's Theorem on $\gamma_1, \gamma_2, \Omega$, we know that $e(\gamma_1, \gamma_2)$ lies on $MN.$ Hence, since it clearly lies on $KL$ (common tangent of $\gamma_1, \gamma_2$), we have that $MN \cap KL=e(\gamma_1, \gamma_2).$ From the three previous Monge applications, it suffices only to show that $e(\gamma_1, \omega), e(\gamma_2, \omega), $ and $e(\gamma_1, \gamma_2)$ are collinear. However, this follows from Monge's Theorem on $\gamma_1, \gamma_2, \omega$, and so we're done! $\square$

AD cuts BC at P, (D,DP) cuts (O) at F,G, denote (O1), (O2) be 2 circle, S=BB-CC so P is incenter of AFG but FG//BC so ALK=AGF=2PGL so PL=LG so LD is bisector of PLG similar for PKF hence AF=AK,AG=AL and D is excenter. LO1 cuts KO2 at Q then Q is incenter of AKL, let distance be r. ML cuts OQ at R then RO/RQ=MO/MO11.LO1/LQ =R/O1P1.O1P1/r=R/r. Similarly we have ML,KN collinear. Notice SO/SD.AD/AQ=SD'/SE.PD/PQ=R/DE.DE/r=R/r=RO/RQ so A,R,S collinear so AR,AE isogonal.

Copied from here for storage purposes. Let $\gamma$ denote the incircle of $\Delta CKL$ Let $T$ be the intersection point of tangents to $\Omega$ at $B$ and $C$. Claim : $\omega$ is the incircle of $\Delta TAB$. Proof: This is just straightforward angle chasing . Just note that $\measuredangle{DAB}=\frac{1}{2}\measuredangle{TAB}$. Now we seek to prove that $CE$ and $CP$ are isogonal in $\measuredangle{ACB}$ . That would clearly finish as we already have that the pairs $\{CA,CB\}$ and $\{CK ,CL\}$ are isogonal in $\measuredangle{ACB}$. Hence we need to prove that $C,P,T$ are collinear . Claim : $P$ is the exsimilicenter of $\gamma$ and $\Omega$ Proof: We use phantom points , and let $P'$ denote the desired exsimilicenter . By Monge on the circles $\{ \Omega ,\Omega_1 ,\gamma \}$ and $\{ \Omega ,\Omega_2 ,\gamma \}$ , we have$$P' \equiv LM \cap KN \implies P' \equiv P .$$To finish , note that by Monge on $\{ \Omega ,\omega ,\gamma \}$ , we get that $C,P,T$ are collinear, as desired . $\blacksquare$