Let $ABCD$ be a cyclic quadrilateral such that $AD=BD=AC$. A point $P$ moves along the circumcircle $\omega$ of triangle $ABCD$. The lined $AP$ and $DP$ meet the lines $CD$ and $AB$ at points $E$ and $F$ respectively. The lines $BE$ and $CF$ meet point $Q$. Find the locus of $Q$.
1. **Setup and Definitions:**
   - Let \(ABCD\) be a cyclic quadrilateral with \(AD = BD = AC\).
   - Let \(P\) be a point moving along the circumcircle \(\omega\) of \(ABCD\).
   - Define points \(E\) and \(F\) as the intersections of lines \(AP\) and \(DP\) with lines \(CD\) and \(AB\), respectively.
   - Define point \(Q\) as the intersection of lines \(BE\) and \(CF\).
   - Let \(X\) be the intersection of diagonals \(AC\) and \(BD\).

2. **Claim 1: \(BX\) is tangent to \(\odot(BPF)\) and \(CX\) is tangent to \(\odot(CPE)\).**
   - **Proof:**
     - Consider the angle \(\angle XBF\). Since \(X\) lies on \(AC\) and \(BD\), we have:
       \[
       \angle XBF = \angle XBA = \angle BAD = \angle BPD
       \]
     - This implies that \(BX\) is tangent to \(\odot(BPF)\) at \(B\).
     - Similarly, for \(\angle XCE\):
       \[
       \angle XCE = \angle XCA = \angle CAD = \angle CPD
       \]
     - This implies that \(CX\) is tangent to \(\odot(CPE)\) at \(C\).
     - Thus, \(BX\) is tangent to \(\odot(BPF)\) and \(CX\) is tangent to \(\odot(CPE)\). \(\blacksquare\)

3. **Claim 2: \(C, Q', F\) and \(B, Q', E\) are collinear, or \(Q' \equiv Q\).**
   - **Proof:**
     - Let \(Q'\) be the second intersection of \(\odot(BPF)\) and \(\odot(CPE)\).
     - From Claim 1, \(PX\) is the radical axis of the two circles, so \(Q'\) lies on \(PX\).
     - Consider the angles:
       \[
       \angle FQ'P = \angle FBP = \angle ACP = \angle XCP = \angle CEP = \angle CQ'P
       \]
     - This shows that \(C, Q', F\) are collinear.
     - Similarly, we can show that \(B, Q', E\) are collinear.
     - Therefore, \(Q' \equiv Q\). \(\blacksquare\)

4. **Conclusion:**
   - From the above claims, \(\odot(BQPF)\) is tangent to \(BX\), so:
     \[
     XB^2 = XQ \cdot XP
     \]
   - This implies that \(Q\) is the image of \(P\) after inversion about \(X\) with radius \(XB^2\).
   - As \(P\) varies on a circle, \(Q\) also varies on a circle which is the image obtained after inverting \(\omega\) about \(X\) with radius \(XB^2\).

The final answer is the locus of \( \boxed{ Q } \) is a circle.