This problem was proposed by Burii.

0 IQ solution Set up a coordinate plane and let $D(0,0)$, $C(a,0)$ and $A(0,b)$. Also let $M(0, \ell)$. We have $\frac{DK}{KC}=\frac{AD}{AC}=\frac{b}{\sqrt{a^2+b^2}}$, hence $DK=\frac{ab}{b+\sqrt{a^2+b^2}}$, implying that $K(\frac{ab}{b+\sqrt{a^2+b^2}},0)$. In addition, $\lambda_{AK}=\frac{-(b+\sqrt{a^2+b^2})}{a}$, and since lines $AK, DL$ are perpendicular, $\lambda_{DL}=\frac{a}{b+\sqrt{a^2+b^2}}$. Since line $DL$ passes through $D(0,0)$, we infer that it is line $y=\frac{a}{b+\sqrt{a^2+b^2}}x$. Line $AC$ passes through points $A(0,b)$ and $C(a,0)$, so it's line $y=-\frac{b}{a}x+b$. Equating lines $AC, DL$ we find that their point of intersection, which is $L$ has the coordinates $L(\frac{ab(b+\sqrt{a^2+b^2}}{a^2+b(b+\sqrt{a^2+b^2})},\frac{a^2b}{a^2+b(b+\sqrt{a^2+b^2})})$. In addition, line $MC$ passes through $M(0,\ell)$ and $C(a,0)$, so it is line $y=-\frac{\ell}{a}x+\ell$, and equating this with line's $DL$ equation we find that $P$ has coordinates $P(\frac{\ell a(b+\sqrt{a^2+b^2})}{a^2+\ell(b+\sqrt{a^2+b^2})},\frac{a^2\ell}{a^2+\ell(b+\sqrt{a^2+b^2})})$. Now, line $ML$ passes through points $M, L$ and we want to find it's intersection with the horizontal axis, which is point $B$. Supposing $L(v,w)$ -I'm not gonna write this mess again - line $ML$ has equation $y=\frac{w-\ell}{v}x+\ell$, so it intersects the horizontal axis at point $B(\frac{\ell v}{\ell-w},0)$, and now putting the values for $v,w$ we find that $B(\frac{\ell ab(b+\sqrt{a^2+b^2})}{\ell(a^2+b(b+\sqrt{a^2+b^2}))-a^2b},0)$. Thus, $\lambda_{AB}=\frac{b}{-\frac{\ell ab(b+\sqrt{a^2+b^2})}{\ell(a^2+b(b+\sqrt{a^2+b^2}))-a^2b}}=\frac{a^2b-\ell(b(b+\sqrt{a^2+b^2})+a^2)}{\ell a(b+\sqrt{a^2+b^2})}$ and $\lambda_{PK}=\frac{\frac{a^2\ell}{a^2+\ell(b+\sqrt{a^2+b^2})}}{\frac{\ell a(b+\sqrt{a^2+b^2})}{a^2+\ell(b+\sqrt{a^2+b^2})}-\frac{ab}{b+\sqrt{a^2+b^2}}}$. Now, by clearing the denominators and simplyfying, it's straightforward to check that $\lambda_{AB} \cdot \lambda_{PK}=-1$, hence $PK \perp AB$, as desired.

sinus Law

Let $PK$ meet $AB$ at $S$. Let $R$ be foot of perpendicular from $L$ to $BC$ and let it meet $BC$ at $Q$. we will prove $QL || PK$. we have $AD = AL$ and $KD = KL$. By Menelaus theorem we have $\frac{LM}{MB} . \frac{DP}{PL} . \frac{BC}{DC} = 1$ and also $\frac{CD}{BD} . \frac{BM}{ML} . \frac{LA}{CA} = 1$ so $\frac{BC}{BD} . \frac{DP}{PL} . \frac{AL}{AC} = 1$. $\frac{BC}{BD} . \frac{DP}{PL} . \frac{AL}{AC} = \sin{ACB} . \frac{DP}{PL} . \frac{BC}{\cos{ABC}} . AB = \frac{DP}{PL} . \frac{\sin{BAC}}{\cos{ABC}} = 1$ so $\frac{DP}{PL} = \frac{\cos {ABC}}{\sin {BAC}}$. we will prove $\frac{DK}{KQ} = \frac{\cos {ABC}}{\sin {BAC}}. \frac{DK}{KQ} = \frac{LK}{KQ} = \frac{\sin {LQK}}{\sin {KLQ}} = \frac{\cos {ABC}}{\sin {BAC}}$.

Let $CP, KP$ cut $AB$ at $S,Q$. By converse of pascal on $ALDDKQ$ it is enough to prove that $SD$ is tangent to $ (AK)$ which comes from the well known fact that $AD$ bisects angle $SDL$.

A longer solution using projective geometry: For projective geometry you usually need lots of new points, so let $k$ be the circle through $A, D, K, L$ and let $F$ be the Intersection of $AB$ and the tangent to $k$ through $D$. Let $X\not= A$ be the second intersection of $k$ and $AB$, let $P'$ be the intersection of $XK$ and $DL$, and $H$ the intersection of $AD$ and $XK$. Let $E$ be the point on $BC$ so that $EA\perp KA$. Now Pascal on $AXKDDL$ gives $F, P', C$ collinear. Let $M'$ be the intersection of this line with $AD$; we need to show that $M'$ lies on $BL$. Pascal on $AAXKDD$ gives $E, F, H$ kollinear. Now (let $L' = BM\cap AC$) \[ -1 = (E, K; D, C) =^H (F, P';M',C) =^B (A, BP\cap AC; L', C) =^P (AP\cap BC, B; L'P\cap BC, C) =^A (P, F; M', C) \]So $L'P$ intersects $BC$ at $D$, so $L'=L, M'=M$ and we are done.

Synthetic solution, anyone ??

Pascal and Blanchet's theorem finishes it