Let $ C_1$ be a circle and $ P$ be a fixed point outside the circle $ C_1$. Quadrilateral $ ABCD$ lies on the circle $ C_1$ such that rays $ AB$ and $ CD$ intersect at $ P$. Let $ E$ be the intersection of $ AC$ and $ BD$. 
(a) Prove that the circumcircle of triangle $ ADE$ and the circumcircle of triangle $ BEC$ pass through a fixed point.
(b) Find the the locus of point $ E$.
Given:
- \( C_1 \) is a circle.
- \( P \) is a fixed point outside the circle \( C_1 \).
- Quadrilateral \( ABCD \) lies on the circle \( C_1 \) such that rays \( AB \) and \( CD \) intersect at \( P \).
- \( E \) is the intersection of \( AC \) and \( BD \).

We need to prove:
(a) The circumcircle of triangle \( ADE \) and the circumcircle of triangle \( BEC \) pass through a fixed point.
(b) Find the locus of point \( E \).

### Part (a)
1. **Construct Tangents and Points:**
   - Let \( PK \) and \( PQ \) be two tangents from \( P \) to \( C_1 \).
   - Let \( O \) be the center of \( C_1 \).
   - Let \( H \) be the intersection of \( PO \) and \( KQ \).
   - Let \( CH \) intersect \( C_1 \) at \( J \).
   - Let \( BH \) intersect \( C_1 \) at \( F \).

2. **Cyclic Quadrilateral:**
   - Since \( HA \cdot HO = HK \cdot HQ = HC \cdot HJ \), it follows that \( PJOC \) is cyclic.
   - Therefore, \( \angle OJH = \angle OPC \).

3. **Similar Triangles:**
   - Since \( OH \cdot OP = R^2 = OD^2 \), we have \( \frac{OH}{OD} = \frac{OD}{OA} \).
   - Thus, \( \triangle HOD \sim \triangle DOP \), implying \( \angle OPC = \angle ODH \).

4. **Perpendicularity and Parallelism:**
   - From the above, \( \angle HJO = \angle HDO \), so \( HO \perp JD \).
   - Therefore, \( JD \parallel KQ \).
   - Similarly, \( AF \parallel KQ \parallel JD \), implying \( \text{arc } AD = \text{arc } JF \).

5. **Cyclic Quadrilaterals:**
   - We have \( \angle BHC = \angle BEC \), so \( BHEC \) is cyclic.
   - Similarly, \( AHED \) is cyclic.

6. **Conclusion:**
   - The circumcircles of triangles \( ADE \) and \( BEC \) pass through a fixed point \( H \).

### Part (b)
1. **Angle Calculation:**
   - We have \( 180^\circ - \angle BHE = \angle BCE = \frac{1}{2} (\text{arc } BK + \text{arc } KA) = \frac{1}{2} (\text{arc } BK + \text{arc } QF) = \angle BHK \).

2. **Collinearity:**
   - Therefore, \( K, H, E \) are collinear.

3. **Locus of \( E \):**
   - Hence, \( E \in [KQ] \).

The final answer is \( \boxed{ E \in [KQ] } \).