a) First, applying Pascal's theorem for 6 points ($DPN$, $MEF$), we get the intersections of pairs of lines ($DE, MP$); ($DF, MN$); ($PF, NE$) collinear or $A$, $X$ and $Y$ are collinear.

Clearly, $180^\circ - \angle BAC = \angle BHC = \angle BKC$ so $K$ lies on $(O)$. Next, we will prove that $XY$ bisects $EF$.

Note that $MN$, $MP$ are the midlines of the triangle $ABC$ so $MN \parallel AB$, $MP \parallel AC$. Therefore, by Thales's theorem, we have
$$\frac{YD}{YF} = \frac{MD}{MB}, \quad XE = \frac{MC}{MD}$$
Let $Q$ be the intersection of $XY$ and $EF$. Applying Menelaus' theorem to triangle $DEF$, we get
$$\frac{QF}{QE} = \frac{XD}{XE}, \quad \frac{YF}{YD} = \frac{MB}{MD}, \quad \frac{MD}{MC} = -1$$
or $Q$ is the midpoint of $EF$. It follows that $AQ$, $AM$ are isogonal with respect to angle $\angle BAC$, so $AQ$ is the symmedian of triangle $ABC$. Therefore, $ABZC$ is a harmonic quadrilateral.

Let $J$ be the intersection of $EF$ and $BC$, then $(JD, BC) = -1$ so $K(JD, BC) = -1$ but we also have $K(ZA, BC) = -1$, which implies $KZ$ passes through $J$.

Finally, we have $JE \cdot JF = JB \cdot JC = JK \cdot JZ$ so $K, Z, E$ and $F$ are concyclic.

b) We have $\angle EBF = \angle ECH = \angle EDH$, $\angle HED = \angle HCD = \angle BEF$ because $BCEF$, $EHDC$ are cyclic quadrilaterals. Therefore, $\triangle BEF \sim \triangle DEH$ (a.a). Hence,
$$\frac{HK}{2EH} = \frac{HD}{EH} = \frac{BF}{EF} = \frac{BF}{2FQ'}$$
but we also have $\angle BFQ = \angle EHK$, then $\triangle BFQ \sim \triangle SLE$ (s.a.s), which implies
$$\angle FBQ = \angle EKH = \angle ABS$$
or $B$, $Q$ and $S$ are collinear. Similarly, $C$, $T$ and $S$ are collinear. Therefore, $BS$, $CT$ and $XY$ are concurrent at $Q$. $\square$