Let $A,B,C,D$, be four different points on a line $\ell$, so that $AB=BC=CD$. In one of the semiplanes determined by the line $\ell$, the points $P$ and $Q$ are chosen in such a way that the triangle $CPQ$ is equilateral with its vertices named clockwise. Let $M$ and $N$ be two points of the plane be such that the triangles $MAP$ and $NQD$ are equilateral (the vertices are also named clockwise). Find the angle $\angle MBN$.
1. **Assigning Complex Numbers:**
   We start by assigning complex numbers to the points on the line $\ell$. Let $C$ be the origin, i.e., $C = 0$. Since $AB = BC = CD$, we can set:
   \[
   A = -2, \quad B = -1, \quad C = 0, \quad D = 1
   \]
   Let $P = p$ be a complex number representing the point $P$.

2. **Using the Sixth Root of Unity:**
   Let $\omega = e^{i\pi/3}$ be the sixth root of unity, which represents a rotation by $60^\circ$ counterclockwise. To rotate a complex number $z$ by $60^\circ$ about the origin, we multiply by $\omega$. To rotate $z$ by $60^\circ$ about another point $z_0$, we use the transformation:
   \[
   z \mapsto (z - z_0)\omega + z_0
   \]

3. **Finding $Q$:**
   Since $\triangle CPQ$ is equilateral with vertices named clockwise, $Q$ is obtained by rotating $P$ by $60^\circ$ counterclockwise about $C$:
   \[
   Q = p\omega
   \]

4. **Finding $M$:**
   The point $M$ is such that $\triangle MAP$ is equilateral. To find $M$, we rotate $A$ by $60^\circ$ counterclockwise about $P$:
   \[
   M = (A - P)\omega + P = (-2 - p)\omega + p = -2\omega - p\omega + p
   \]

5. **Finding $N$:**
   The point $N$ is such that $\triangle NQD$ is equilateral. To find $N$, we rotate $D$ by $60^\circ$ counterclockwise about $Q$:
   \[
   N = (D - Q)\omega + Q = (1 - p\omega)\omega + p\omega = \omega - p\omega^2 + p\omega
   \]
   Since $\omega^2 = -\omega + 1$, we have:
   \[
   N = \omega - p(-\omega + 1) + p\omega = \omega + p\omega - p + p\omega = \omega + 2p\omega - p
   \]

6. **Simplifying $M$ and $N$:**
   We simplify the expressions for $M$ and $N$:
   \[
   M = -2\omega - p\omega + p
   \]
   \[
   N = \omega + 2p\omega - p
   \]

7. **Redefining $B$ as the Origin:**
   To simplify the calculation of $\angle MBN$, we redefine $B$ as the origin. Thus, we shift all points by adding $1$:
   \[
   M' = M + 1 = -2\omega - p\omega + p + 1
   \]
   \[
   N' = N + 1 = \omega + 2p\omega - p + 1
   \]

8. **Verifying Rotation:**
   We need to verify that $N'$ is $M'$ rotated by $60^\circ$ about $B$:
   \[
   M'\omega = (-2\omega - p\omega + p + 1)\omega = -2\omega^2 - p\omega^2 + p\omega + \omega
   \]
   Using $\omega^2 = -\omega + 1$:
   \[
   M'\omega = -2(-\omega + 1) - p(-\omega + 1) + p\omega + \omega = 2\omega - 2 + p\omega - p + p\omega + \omega
   \]
   Simplifying:
   \[
   M'\omega = 3\omega - 2 - p + 2p\omega
   \]
   Comparing with $N'$:
   \[
   N' = \omega + 2p\omega - p + 1
   \]
   We see that $M'\omega = N'$, confirming that $N'$ is indeed $M'$ rotated by $60^\circ$ about $B$.

Conclusion:
Since $N'$ is $M'$ rotated by $60^\circ$ about $B$, the angle $\angle MBN$ is $60^\circ$.

The final answer is $\boxed{60^\circ}$.