Incircle Excircle problem of a triangle
Observe that, $\triangle PAM\cong \triangle LCP$ since $PC=AK=AM$($\angle AMK=\angle AKM$ through angle chasing), $AP=KC=LC$ and $\angle MAP= \angle PCL$. Hence, $PL=PM$.
Observe that, $\triangle PAM\cong \triangle LCP$ since $PC=AK=AM$($\angle AMK=\angle AKM$ through angle chasing), $AP=KC=LC$ and $\angle MAP= \angle PCL$. Hence, $PL=PM$.