Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\Omega$ with $AC \perp BD$. Let $P=AC \cap BD$ and $W,X,Y,Z$ be the projections of $P$ on the lines $AB, BC, CD, DA$ respectively. Let $E,F,G,H$ be the mid-points of sides $AB, BC, CD, DA$ respectively.

(a) Prove that $E,F,G,H,W,X,Y,Z$ are concyclic.

(b) If $R$ is the radius of $\Omega$ and $d$ is the distance between its centre and $P$, then find the radius of the circle in (a) in terms of $R$ and $d$. 
### Part (a)

1. **Given Setup**: 
   - Let $ABCD$ be a cyclic quadrilateral inscribed in circle $\Omega$ with $AC \perp BD$.
   - Let $P = AC \cap BD$.
   - Let $W, X, Y, Z$ be the projections of $P$ on the lines $AB, BC, CD, DA$ respectively.
   - Let $E, F, G, H$ be the mid-points of sides $AB, BC, CD, DA$ respectively.

2. **Prove Concyclicity**:
   - By Brahmagupta's Theorem, the perpendiculars from $P$ to the sides of the cyclic quadrilateral $ABCD$ are concurrent at the orthocenter of the quadrilateral.
   - Since $AC \perp BD$, $P$ is the orthocenter of the quadrilateral $ABCD$.
   - The projections $W, X, Y, Z$ of $P$ on $AB, BC, CD, DA$ respectively form a rectangle because $P$ is the orthocenter.
   - The midpoints $E, F, G, H$ of the sides $AB, BC, CD, DA$ respectively form another rectangle because the diagonals of a cyclic quadrilateral bisect each other.
   - Since $E, F, G, H$ form a rectangle and $W, X, Y, Z$ form a rectangle, and both rectangles share the same center (the midpoint of $P$), they are concyclic.

3. **Angle Chasing**:
   - $\angle EWG = 90^\circ$ because $W$ is the foot of the perpendicular from $P$ to $AB$ and $E$ is the midpoint of $AB$.
   - Similarly, $\angle FXH = 90^\circ$, $\angle GYF = 90^\circ$, and $\angle HZP = 90^\circ$.
   - Therefore, $E, F, G, H, W, X, Y, Z$ lie on the same circle.

Thus, $E, F, G, H, W, X, Y, Z$ are concyclic.

### Part (b)

1. **Radius Calculation**:
   - The radius of the circle passing through $E, F, G, H, W, X, Y, Z$ is the same as the circumradius of the rectangle $EFGH$.
   - The diagonal of the rectangle $EFGH$ is $\sqrt{AC^2 + BD^2}$.
   - Since $AC \perp BD$, $AC$ and $BD$ are the diameters of the circle $\Omega$.

2. **Using Pythagoras Theorem**:
   - Let $O$ be the center of $\Omega$ and $R$ be its radius.
   - The distance $d$ between $O$ and $P$ is given.
   - Drop perpendiculars from $O$ to $AC$ and $BD$ at points $S$ and $T$ respectively.
   - Since $OSPT$ is a rectangle, $OS = OT = R$.
   - By Pythagoras' theorem in $\triangle OSP$ and $\triangle OTP$, we have:
     \[
     OP^2 = OS^2 + SP^2 = R^2 + d^2
     \]
   - Therefore, the radius of the circle in part (a) is:
     \[
     \text{Radius} = \frac{1}{2} \sqrt{AC^2 + BD^2} = \frac{1}{2} \sqrt{2R^2 - d^2}
     \]

The final answer is $\boxed{\frac{\sqrt{2R^2 - d^2}}{2}}$