a) It is well known that $KE$, $KF$ are both tangent to $(I)$. Thus, $GK$ is the symmedian of $\angle GEF$, it follows that $\overarc{LE} = \overarc{HF}$. Hence, $AH$, $AL$ are isogonal with respect to angle $BAC$. It is clear that $AH$ is the diameter of $(I)$. Therefore $AL$ is the altitude of $\angle AEF$.

b) Since $I$ is the midpoint of $AH$, it's clear that $(IEF)$ is the Euler's circle of $\angle ABC$ with the diameter $IK$. Besides,
$$\overline{MI} \cdot \overline{MN} = \overline{ME} \cdot \overline{MF} = \overline{MA} \cdot \overline{ML},$$
this implies that $A$, $I$, $L$ and $N$ are concyclic. Therefore,
$$\angle ANI = \angle ALI = \angle LAI = \angle DIK,$$
since $IK \parallel AL$ (both lines are perpendicular to $EF$). Hence,
$$\angle AND = \angle ANI + \angle IND = \angle DIK + \angle IKD = 90^\circ.$$
Let $S$ be the radical center of $(I)$, $(IEF)$ and $(ADN)$. Since $EF$ is the radical axis of $(I)$ and $(IEF)$ then $EF$ passes through $S$. Similarly, $DN$ passes through $S$. Since the centers of $(AND)$, $I$ and $A$ are collinear, we have $(AND)$ and $(I)$ are tangent at $A$, thus $AS$ is tangent to $(I)$, in other words, $AS \parallel BC$. Hence, $A(SK, QP) = A(SK, CB) = -1$, it follows that $PE$, $QF$ and $AK$ are concurrent. $\square$