Given a circle, point $C$ on it and point $A$ outside the circle. The equilateral triangle $ACP$ is constructed on the segment $AC$. Point $C$ moves along the circle. What trajectory will the point $P$ describe?
1. **Define the given elements:**
   - Let the circle have center \( O(0,0) \) and radius \( r \).
   - Let point \( A \) be outside the circle with coordinates \( A(a, b) \).
   - Point \( C \) is on the circle, so its coordinates can be parameterized as \( C(r \cos \theta, r \sin \theta) \) where \( \theta \) is the angle parameter.

2. **Construct the equilateral triangle \( ACP \):**
   - Since \( ACP \) is an equilateral triangle, the length of \( AC \) is equal to the length of \( CP \) and \( AP \).
   - The coordinates of point \( P \) can be found using rotation. If we rotate point \( C \) by \( 60^\circ \) counterclockwise around point \( A \), we get the coordinates of \( P \).

3. **Calculate the coordinates of \( P \):**
   - The rotation matrix for \( 60^\circ \) counterclockwise is:
     \[
     R_{60^\circ} = \begin{pmatrix}
     \cos 60^\circ & -\sin 60^\circ \\
     \sin 60^\circ & \cos 60^\circ
     \end{pmatrix} = \begin{pmatrix}
     \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
     \frac{\sqrt{3}}{2} & \frac{1}{2}
     \end{pmatrix}
     \]
   - The coordinates of \( P \) after rotating \( C \) around \( A \) are:
     \[
     P = A + R_{60^\circ} \cdot (C - A)
     \]
   - Substituting \( C = (r \cos \theta, r \sin \theta) \) and \( A = (a, b) \):
     \[
     P = (a, b) + \begin{pmatrix}
     \frac{1}{2} & -\frac{\sqrt{3}}{2} \\
     \frac{\sqrt{3}}{2} & \frac{1}{2}
     \end{pmatrix} \cdot \begin{pmatrix}
     r \cos \theta - a \\
     r \sin \theta - b
     \end{pmatrix}
     \]
   - Performing the matrix multiplication:
     \[
     P = (a, b) + \begin{pmatrix}
     \frac{1}{2}(r \cos \theta - a) - \frac{\sqrt{3}}{2}(r \sin \theta - b) \\
     \frac{\sqrt{3}}{2}(r \cos \theta - a) + \frac{1}{2}(r \sin \theta - b)
     \end{pmatrix}
     \]
     \[
     P = \left( a + \frac{1}{2}(r \cos \theta - a) - \frac{\sqrt{3}}{2}(r \sin \theta - b), b + \frac{\sqrt{3}}{2}(r \cos \theta - a) + \frac{1}{2}(r \sin \theta - b) \right)
     \]

4. **Simplify the coordinates of \( P \):**
   - Let \( x = r \cos \theta \) and \( y = r \sin \theta \):
     \[
     P = \left( a + \frac{1}{2}(x - a) - \frac{\sqrt{3}}{2}(y - b), b + \frac{\sqrt{3}}{2}(x - a) + \frac{1}{2}(y - b) \right)
     \]
     \[
     P = \left( \frac{a + x}{2} - \frac{\sqrt{3}}{2}(y - b), \frac{b + y}{2} + \frac{\sqrt{3}}{2}(x - a) \right)
     \]

5. **Determine the trajectory of \( P \):**
   - As \( C \) moves around the circle, \( x \) and \( y \) trace out the circle \( x^2 + y^2 = r^2 \).
   - The coordinates of \( P \) are linear combinations of \( x \) and \( y \) with constant coefficients, which means \( P \) will trace out a circle as well.
   - The center of this circle can be found by averaging the transformations applied to \( x \) and \( y \).

6. **Find the center and radius of the trajectory circle:**
   - The center of the trajectory circle is the midpoint of the transformation:
     \[
     M = \left( \frac{a - \sqrt{3} b}{2}, \frac{\sqrt{3} a + b}{2} \right)
     \]
   - The radius of the trajectory circle is the same as the original circle, \( r \).

7. **Conclusion:**
   - The trajectory of point \( P \) is a circle with center \( M \) and radius \( r \).

The final answer is \(\boxed{\text{The trajectory of point } P \text{ is a circle with center } \left( \frac{a - \sqrt{3} b}{2}, \frac{\sqrt{3} a + b}{2} \right) \text{ and radius } r.}\)