I think there is some mistake in the wording since this is not necessarily true always, can you kindly check it?

No I think the problem is fine. This is just EGMO Lemma 4.17

I made a little mistake, I now corrected it. The order was supposed to be "tangent to the sides $AB,AC$ and $BC$ at the points $M,N$ and $K$" but I wrote "tangent to the sides $AB,BC$ and $CA$ at the points $M,N$ and $K$"

Let the line parallel to $BC$ through $A$ meet $MN$ at $X$. $L$ lies on the polar of $A$, so $A$ lies on the polar of $L$. Also, $(M,N;L,X)=(B,C;D,\infty)=-1$, so $X$ lies on the polar of $L$. Thus the polar of $L$ is $AX$, which intersects the polar of $K$ ($BC$) at infinity, i.e. on the polar of $I$. Hence $K,I,L$ are collinear.

Wait how did they directly put Lemma $4.17$ in a TST that even P3???

Well it's problem $3$ out of $5$ from a JBMO TST. I guess it is not that well known, is it?

Let the line parallel to $BC$ passing through $L$ be $M'N'$, and we claim that $L,I,K$ colleniar.Now we'll prove that $L$ is the midpoint of $M'N'$. Since $LK\perp M'N'$ and $IM\perp AB$ , we get $LM'MI$ cyclic. And $IN\perp AC \implies LNN'I$ cyclic.We'll prove that $AM'IN'$ is cyclic.We have $AMIN$ cyclic , so if $\angle{M'IM}= \angle{NIN'}$ $\implies AM'IN'$ cyclic. And this is easy to see since $\angle{M'LM} = \angle{NLN'}$. Now, we have $IM'=IN'$, $\angle IM'N'=\angle IN'M' , IL\perp M'N'$ $\implies M'L=N'L$ and we're done

Let $R$=$MN \cap KI$. Draw a $B'C' \parallel BC$ meeting $AB$ at $B'$ and $AC$ at $C'$. It suffices to show that $R$ is the midpoint of $B'C'$ Claim $1$: $I$ lies on $AB'C'$ Proof: $IM$ and $IN$ are perpendiculars to side $AB'$ and $AC'$. Also $IR \perp B'C'$ so $MN$ is the Simson line of $\triangle AB'C'$ w.r.t $I$. So, $I$ must lie on the circumcircle of $AB'C'$. Now, due to Incentre Excentre Lemma (Fact-5), We have $IB'=IC'$ which gives, $\triangle IB'R \cong \triangle IC'R$. Thus $R$ is the midpoint of $B'C'$. Our proof is thus complete.