a. Note that $\angle DEB = \angle DFC$ then $\angle DEA = \angle DFA$, which implies $MNEF$ is a cyclic quadrilateral.

We have $\angle DI_2F = 2\angle DNF = 2\angle EMF$ and
$$\angle I_2DF = 90^\circ - \frac{1}{2} \angle DI_2F$$
so $I_2D \perp ME$. We also have $J_1K \perp ME$ and it follows that $I_2D \parallel J_1K$. Similarly, we get $I_1D \parallel J_2H$. Hence, $\angle I_2DK = \angle DKJ_1 = \angle KDJ_1$ or $DK$ is the bisector of $\angle I_2DI_1$. Similarly, we also have $DH$ is the bisector of $\angle I_2DI_1$. Hence, three points $D$, $H$ and $K$ are collinear.

Since $MNEF$ is cyclic, $AE \cdot AM = AF \cdot AN$, so $A$ belongs to the radical axis of $(I_1)$ and $(I_2)$ which implies $A$, $D$ and $P$ are collinear. Furthermore,
$$\angle AKH = 90^\circ - \angle DKJ_1 = 90^\circ - \angle DHJ_2 = \angle DHF = \angle AHK$$
so $AH = AK$. It follows that $A$ has the same power to $(J_1)$ and $(J_2)$, or $A$ lies on the radical axis of $(J_1)$ and $(J_2)$. Hence, $A$, $D$ and $Q$ are collinear.

From these, we have four points $A$, $D$, $P$, $Q$ are collinear.

b. Since $AK$ is a tangent of $(J_1)$ then $\angle AQK = \angle AKD = \angle AHK$, it follows that $AQHK$ is cyclic. We have $\angle GEF = \angle GAF = \angle GKH$, $\angle GHK = \angle GAK = \angle GFE$ so $\triangle GEF \sim \triangle GKH$ (a.a).

Take the point $S$ in $EF$ such that $\overline{SE} : \overline{DF} = \overline{DK} : \overline{DH}$, then $\triangle GES \sim \triangle GKD$ (s.a.s). Thus, $\triangle GEK \sim \triangle GSD$ (s.a.s). From here, we have
$$\angle GDS = \angle GKE = \angle GQD$$
so $DS$ is the tangent of the circumcircle of triangle $GDQ$. Similarly, $\triangle LEF \sim \triangle QKH$ (a.a) so $\triangle LES \sim \triangle QKD$ (s.a.s). Hence, $\angle KQD = \angle ELS$, and
$$\angle KQG = \angle KHG = \angle EFG = \angle ELG$$
which implies $\angle SLG = \angle DQG = \angle GDS$. It shows that $DLGS$ is a cyclic quadrilateral. From these results, the problem is proved. $\square$