In the interior of the square $ABCD$, the point $P$ lies in such a way that $\angle DCP = \angle CAP=25^{\circ}$. Find all possible values of $\angle PBA$.
1. **Reflecting Point \( C \) Across \( B \)**:
   - Let \( C' \) be the reflection of \( C \) across \( B \). Since \( B \) is the midpoint of \( CC' \), we have \( BC = BC' \).

2. **Cyclic Quadrilateral \( APCC' \)**:
   - Since \( \angle DCP = \angle CAP = 25^\circ \), we need to show that \( APCC' \) is a cyclic quadrilateral.
   - By reflecting \( C \) to \( C' \), we ensure that \( \angle DCP = \angle C'CP \) and \( \angle CAP = \angle C'AP \).

3. **Circle Centered at \( B \)**:
   - Since \( C' \) is the reflection of \( C \) across \( B \), \( B \) is equidistant from \( C \) and \( C' \). Therefore, \( B \) is the center of the circle passing through \( A, P, C, \) and \( C' \).

4. **Isosceles Triangle \( \triangle BAP \)**:
   - Since \( B \) is the center of the circle, \( BA = BP = BC \). This implies that \( \triangle BAP \) is isosceles with \( BA = BP \).

5. **Calculating \( \angle PBA \)**:
   - In the cyclic quadrilateral \( APCC' \), the opposite angles sum up to \( 180^\circ \). Therefore, \( \angle APC + \angle ACC' = 180^\circ \).
   - Since \( \angle DCP = \angle CAP = 25^\circ \), we have \( \angle APC = 180^\circ - 2 \times 25^\circ = 130^\circ \).
   - In \( \triangle BAP \), since \( BA = BP \), the base angles are equal. Let \( \angle PBA = x \). Then, \( \angle BAP = x \).
   - The sum of angles in \( \triangle BAP \) is \( 180^\circ \), so \( x + x + 130^\circ = 180^\circ \).
   - Solving for \( x \), we get:
     \[
     2x + 130^\circ = 180^\circ
     \]
     \[
     2x = 50^\circ
     \]
     \[
     x = 25^\circ
     \]

6. **Conclusion**:
   - Therefore, the possible value of \( \angle PBA \) is \( 25^\circ \).

The final answer is \( \boxed{25^\circ} \).