a) It is easy to see that $K$ belongs to segment $AC$. Since $IK = IC$, we have triangle $IKC$ is isosceles at $I$. Hence, $\angle AKI = 180^\circ - \angle IKC = 180^\circ - \angle ICK = \angle ABI$, because $ABIC$ is a cyclic quadrilateral. Furthermore, note that $I$ is the midpoint of arc $BC$, then $\angle IAK = \angle IAB$ and $IK = IB = IC$ so $\triangle ABI = \triangle AKI$.

We conclude that $AI$ is the perpendicular bisector of the segment $BK$, which implies that $E$ is the midpoint of $BK$. Note that,
$$\angle DCK = \angle ABD = \angle AKB = \angle DKC$$
so triangle $DKC$ is isosceles, hence $DK = DC$. Note that $IK = IC$, we can see that $ID$ is also the perpendicular bisector of $KC$ so $F$ is the midpoint of $CK$. From these above results, we have $EF$ is the midline of triangle $KBC$, then $EF = \frac{1}{2}BC$.

b) Let $J$ be the midpoint of arc $BC$ of $(O)$ that contains $A$, it is well-known that $IJ$ is the diameter of circle $(O)$. We will show that $J$, $K$ and $P$ are collinear.

Indeed, from $\angle IPJ = 90^\circ$, we only need to prove $\angle KPI = 90^\circ$. In triangle $ADI$, because $DK \perp AI$, $AK \perp DI$ then $K$ is the orthocenter of triangle $ADI$. Hence, $IK$ is perpendicular to $AD$.

On the other hand, note that $CM \parallel AD$, it follows that $CM \perp IK$, but $IM \perp KC$ so $M$ is the orthocenter of triangle $IKC$. Therefore,
$$\angle MKC = 90^\circ - \angle KCI = 90^\circ - (\angle ACB + \frac{1}{2}\angle BAC), \text{ so}$$
$$\angle KNB = \angle NKC + \angle NCK = 90^\circ - \frac{1}{2}\angle BAC.$$
We have $\angle KPI = \angle KPB + \angle BPI = \angle KNB + \angle IAB = 90^\circ - \frac{1}{2}\angle BAC + \frac{1}{2}\angle BAC = 90^\circ$. Hence, $J$, $P$ and $K$ are collinear.

We also have $AJ \perp AI$, $KD \perp AI$ so $AJ \parallel KD$ and $DJ \perp ID$, $AK \perp ID$ then $DJ \parallel AK$. Therefore, $AJDK$ is a parallelogram and the line $PK$ bisects the segment $AD$.