I think there is some typo.

I think $G$ should be the centroid. Then the problem is fine and here is the solution. Firstly, note that $\angle BPK = \angle LNC = \alpha$. Also, we have that $\angle PKB = \angle PGB = \angle NGC = \angle NLC.$ From here it follows that $\triangle NCL \sim \triangle PBK$. In particular, we have that $\angle PBK = \angle NCL$. Claim: $\triangle ACL \sim \triangle ABK$. Proof: Since $\angle ACL = \angle ABK$, it is enough to show that $\frac{AC}{CL}=\frac{AB}{BK}\iff \frac{AC}{AB}=\frac{CL}{BK}$. However, we know that $\frac{NC}{CL}=\frac{PB}{BK}$. Hence, $$\frac{CL}{BK}=\frac{NC}{PB}=\frac{2NC}{2PB}=\frac{AC}{AB},$$so we are done. From the claim it follows that $\angle CAL=\angle BAK$ and we are done.

$G$ is centroid and not incenter as I had written at the start. It has been corrected. Thanks for noticing.

Let $Q=(BPG)\cap (CGN)$, it s clear that $Q$ is the Miquel point of $CNBP$ then $QCN\sim QPB$ so the ratio of the respective $Q$-heights is equal to $\frac{CN}{PB} =\frac{CA}{AB}$ which means that $Q$ is on the $A$-symmedian of $ABC$ more since $Q$ is the Miquel point of $CNBP$ we deduces that $APQC,BQNA$ are cyclic thus $\angle KQG=\angle KNG=\angle MNG=\angle PBG=\angle PQG$ so $P,K,Q$ are collinear similarly we get $N,L,Q$ are collinear hence : since $(AN,AP)$ and $(AM,AQ)$ are 2 pairs of isogonals wrt $\angle BAC$ and $PQ\cap MN=L, NQ\cap MP=K$ then by the isogonal property we deduce $AK,AL$ are also isogonal. Best regards. RH HAS