It's easy to see that $T$ is nothing but the $C$-intouch point. Redefine $P$ as the second intersection of $LT$ with $k$. Then it suffices to show that $MP,AC,IT$ are concurrent (where $I$ is incenter of $\triangle ABC$), since the perpendicular from $T$ to $AB$ passes through $I$. By Pascal on $CMBAPL$, we get that $CM \cap AP,BM \cap PL=IM \cap PT$ and $AB \cap CL=AT \cap CI$ are collinear. This means that $\triangle MIC$ and $\triangle PTA$ are perspective, which directly gives that $MP,IT,AC$ are concurrent. REMARK: It's well known that $P$ is in fact the second intersection of the circle with diameter $AI$ and $k$. However, I don't think that this helps in a shorter proof of the given problem. P.S. The problem given above has a typo in the second statement. The perpendicular should be from $T$ and not $P$.