[asy][asy] import graph; size(17.500557172627527cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-8.860535490411433,xmax=8.640021682216094,ymin=-4.551851715943749,ymax=8.845126533447038; pair A=(-1.461003513516025,6.852595272167005), B=(-2.1382840950205244,-0.2995733029363512), C=(8.024804307410092,-0.272968359474491), I=(0.5968442371705093,2.195336715973298), F=(-1.8798175285459484,2.429866319188775), D=(0.6033566270892435,-0.2923962329831843), K=(3.4223196567412906,4.772658902103807), L=(-1.119683834720336,2.7657375948057177), M=(-8.092277176050304,-0.3151596722584186), Q=(0.5933695578523076,3.52266421552634), P=(6.288939985867701,1.630030782934962); draw(circle(I,2.487741473012453),linewidth(2.)); draw(A--B,linewidth(2.)); draw(B--C,linewidth(2.)); draw(C--A,linewidth(2.)); draw(M--K,linewidth(2.)); draw(M--B,linewidth(2.)); draw(B--K,linewidth(2.)); draw(K--C,linewidth(2.)); draw(B--I,linewidth(2.)); draw(C--I,linewidth(2.)); draw(I--L,linewidth(2.)); draw(Q--D,linewidth(2.)); draw(P--Q,linewidth(2.)); draw(circle((4.310824243124892,0.9611841781730541),3.9136657970741062),linewidth(2.)+linetype("4 4")); draw(K--D,linewidth(2.)); draw(M--P,linewidth(2.)); dot(A,linewidth(3.pt)+ds); label("$A$",(-1.347365255955822,7.034724067313149),NE*lsf); dot(B,linewidth(3.pt)+ds); label("$B$",(-2.2223931145871982,-0.9008734092404037),NE*lsf); dot(C,linewidth(3.pt)+ds); label("$C$",(8.006380819069236,-0.8707000348048389),NE*lsf); dot(I,linewidth(3.pt)+ds); label("$I$",(0.7345975800981424,2.5388912764139877),NE*lsf); dot(F,linewidth(3.pt)+ds); label("$F$",(-2.3430866123294574,2.629411399720682),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(0.2819969635646718,-0.8707000348048389),NE*lsf); dot((2.090992783868283,4.184401259156122),linewidth(3.pt)+ds); label("$E$",(2.062226055262989,4.530333989161267),NE*lsf); dot(K,linewidth(3.pt)+ds); label("$K$",(3.631241525912354,5.163974852308129),NE*lsf); dot(L,linewidth(3.pt)+ds); label("$L$",(-1.1964983837779986,3.232878888431979),NE*lsf); dot(M,linewidth(3.pt)+ds); label("$M$",(-8.287241376135704,-0.840526660369274),NE*lsf); dot(Q,linewidth(3.pt)+ds); label("$Q$",(0.31217033800023647,3.8966931260144055),NE*lsf); dot((3.452560236344878,1.2463846176656133),linewidth(4.pt)+ds); label("$N$",(3.5708947770412243,1.4828231711692184),NE*lsf); dot(P,linewidth(4.pt)+ds); label("$P$",(6.407191973984307,1.8750770388315612),NE*lsf); label("$d$",(2.3337864251830713,3.6854795049654516),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] (a) We angle chase to show that $\angle{BKC}=90^{\circ}.$ It suffices to prove that $K$ is on the circumcircle of cyclic quadrilateral $DCEI$. Indeed, bearing in mind that $K$ is on the $\angle{B}$ internal bisector and thus equidistant from $F$ and $D$, \[ \angle{KDC} = 180^{\circ} - \angle{BDK} = 180^{\circ} - \angle{BFK} = \angle{AFE} = \angle{AEF} = \angle{KEC} . \]Hence $K$ lies on the circumcircle of points, $C, D, E$. This proves the lemma. (b) Let $N$ be the midpoint of $CL$. It well known that $-1 = (BC; MD)$. Then \[ -1 = (BC ; MD) \overset{I}{=} (KL; MQ) \overset{P}{=} (LC, N, PQ \cap CL) . \]Hence $PQ \parallel CL$ since $N$ is the midpoint of $CL$.

$(a)$ Its well known that this is $90^{\circ}$ $(b)$Define $G=MP\cap CI,H=PQ\cap KG$. So $(M,D,B,C)\overset{L}{=} (M,Q,K,L)\overset{G}{=} (P,Q,H,PQ\cap CI)$.Thus it suffices to show that $G$ is the midpoint of $CL$.Maybe I am just dumb but I couldn't find any synthetic for proving this Lol Now I see that this is the definition of $G$ only

You know what's coming. I hope synthetic mafia won't send their mobster. Ahh, I don't care. I'm like Ace Rothstein in this Casino. Anyway, I can hide behind the imaginary axis so they won't get me. Let $x,y,z$ be complex numbers lying on unit circle centered at zero (coordinate of incenter), such that they represent points $D,E,F$ respectively on Argand plane. By already existing formulas $$b=\frac{2xz}{x+z}, \ c=\frac{2xy}{x+y},\ m=\frac{x^2(y+z)-2xyz}{x^2-yz}.$$$$K\in BI\iff \frac{b}{k}=\overline{\left(\frac{b}{k}\right)}\iff\overline{k}=\frac{k}{xz},\ K\in EF\iff \overline{k}=\frac{y+z-k}{yz}$$Hence we get coordinates of $K$ and by swapping variables $y,z$ a formula for $l$ too $$k=\frac{x(y+z)}{x+y},\ l=\frac{x(y+z)}{x+z}.$$(A) We will prove $BK\perp CK$. This is equivalent with $$\frac{c-k}{b-k}=-\overline{\left(\frac{c-k}{b-k}\right)}.$$Indeed$$\frac{c-k}{b-k}=\frac{z-x}{x+z}=-\overline{\left(\frac{z-x}{x+z}\right)}$$because $$b-k=x\cdot\frac{(y-z)(z-x)}{(x+y)(x+z)},\ c-k=\frac{x(y-z)}{x+y}.$$(B)$$Q\in DI\iff \frac{q}{x}=\overline{\left(\frac{q}{x}\right)}\iff\overline{q}=\frac{q}{x^2},\ Q\in EF\iff \overline{q}=\frac{y+z-q}{yz}$$Therefore $$q=\frac{x^2(y+z)}{x^2+yz}$$From midpoint definition $$n=\frac{c+l}{2}=x\cdot\frac{y^2+3y(x+z)+xz}{(x+y)(x+z)}.$$The time's up for $P$. $$P\in CK\iff \frac{c-p}{c-k}=\overline{\left(\frac{c-p}{c-k}\right)}\iff \frac{2xy-p(x+y)}{x(y-z)}=\frac{2z-\overline{p}(x+y)z}{z-y}\iff \overline{p}=\frac{2x(y+z)-p(x+y)}{xz(x+y)}.$$Let's have an auxiliary calculation:$$n-m= \frac{2x(y-z)(x - y) (x^2 + 2 x z + y z) }{(x^2-yz)(x+y)(x+z)}.$$Hence$$\frac{m-p}{n-m}=\frac{x^2(y+z)-2xyz+(yz-x^2)p}{2x(y-z)(x-y)}\cdot \frac{(x+y)(x+z)}{x^2+2xz+yz},\ \overline{\left(\frac{m-p}{n-m}\right)} =\frac{(y+z-2x)xz(x+y)+(x^2-yz)\left(2x(y+z)-p(x+y)\right)}{2xz(y-z)(x-y)}\cdot \frac{y(x+z)}{x^2+2xy+yz}$$En route to equation on the right there is division by $x+y$ on nominator and denominator (just so you can have full factorization if you're still checking ). We have $$P\in MN\iff \frac{m-p}{n-m}= \overline{\left(\frac{m-p}{n-m}\right)}\iff$$It's just a linear equation in $p$. Strike a pose and let's get to it.$$p=\frac{x}{(x+y)(x^2-yz)}\cdot\frac{(x^2+2xz+yz)y(z(x+y)(y+z-2x)+2(y+z)(x^2-yz))-z(xy+xz-2yz)(x+y)(x^2+2xy+yz)}{y(x^2+2xz+yz)-z(x^2+2xy+yz)}$$$$p=\frac{x(y-z)(x^2-yz)\left(x^2(2y+z)+y^2z\right)}{(x+y)(y-z)(x^2-yz)(x^2+yz)}=\frac{x\left(x^2(2y+z)+y^2z\right)}{(x+y)(x^2+yz)}$$And the cherry on top... We will prove $PQ\parallel CI$ as it's easier and equivalent with $PQ\parallel CL$. $$\frac{p-q}{c}=\frac{xy(x-y)(x-z)}{(x+y)(x^2+yz)}\cdot \frac{x+y}{2xy}=\frac{(x-y)(x-z)}{2(x^2+yz)}$$Finally $$\frac{p-q}{c}=\frac{(x-y)(x-z)}{2(x^2+yz)}=\overline{\left(\frac{(x-y)(x-z)}{2(x^2+yz)}\right)}=\overline{\left(\frac{p-q}{c}\right)}\implies PQ\parallel CI.$$QED

Yup, "very well known" seems like an understatement.

Oops my solutions are basically identical to previous ones, but someone sent this to me so I'll post for storage. By the Iran Lemma, we know $\angle BKC = 90^{\circ}$. Furthermore, since $\angle BKC = \angle BLC = 90^{\circ}$, we know $DI, BL, CK$ concur at the orthocenter $H$ of $BIC$. Let $N$ be the midpoint of $CL$. Now, notice $$(C, L; N, CL \cap PQ) \overset{P}{=} (K, L; M, Q) \overset{H}{=} (C, B; M, D) = -1$$where the last equality follows from Ceva-Menelaus on the Gergonne point of $ABC$. $\blacksquare$

Very nice problem.

Nice way to practice projective geo

Let the midpoint of $CL$ be $N$. By Iran Lemma we have $\angle BLC = \angle BKC = 90^\circ$ which implies that $BL$, $DQ$ and $CK$ concur at the orthocenter of $\triangle BIC$. Also, notice that $(B, C; D, M) = -1$ by Ceva-Menelaus. Then projecting from the orthocenter of $\triangle BIC$ we get that $(L, K; Q, M) = -1$, and projecting from $P$ onto $CL$ gives $(L, C; PQ \cap CL, N) = -1$ which implies that $PQ \parallel CL$ as desired.

Let $N$ be the midpoint of $CL$. Notice that \[-1 = (BC;DM) \overset{I}{=} (KL;QM) \overset{P}{=} (CL;PQ \cap CL, N),\] which gives the desired. $\blacksquare$