Quadrangle $ ABCD$ is inscribed in a circle with radius 1 in such a way that the diagonal $ AC$ is a diameter of the circle, while the other diagonal $ BD$ is as long as $ AB$. The diagonals intersect at $ P$. It is known that the length of $ PC$ is $ 2/5$. How long is the side $ CD$?
1. **Identify the given information and draw the circle:**
   - Quadrilateral \(ABCD\) is inscribed in a circle with radius 1.
   - Diagonal \(AC\) is a diameter of the circle.
   - Diagonal \(BD\) is equal in length to side \(AB\).
   - The diagonals intersect at point \(P\).
   - The length of \(PC\) is \(\frac{2}{5}\).

2. **Use the property of the circle:**
   - Since \(AC\) is a diameter, \(\angle ABC = \angle ADC = 90^\circ\) (Thales' theorem).

3. **Apply the given information:**
   - Let \(O\) be the center of the circle. Since \(AC\) is a diameter, \(O\) is the midpoint of \(AC\).
   - Therefore, \(OA = OC = 1\).

4. **Use the given length \(PC = \frac{2}{5}\):**
   - Since \(P\) lies on \(AC\), and \(AC\) is a diameter, \(P\) divides \(AC\) into two segments: \(AP\) and \(PC\).
   - Let \(AP = x\). Then \(PC = 1 - x\).
   - Given \(PC = \frac{2}{5}\), we have \(1 - x = \frac{2}{5}\), so \(x = \frac{3}{5}\).

5. **Determine the coordinates of points:**
   - Place the circle in the coordinate plane with \(O\) at the origin.
   - Let \(A = (-1, 0)\) and \(C = (1, 0)\).
   - Point \(P\) divides \(AC\) in the ratio \(\frac{3}{5}\) to \(\frac{2}{5}\), so \(P = \left(\frac{3}{5} \cdot 1 + \frac{2}{5} \cdot (-1), 0\right) = \left(\frac{3}{5} - \frac{2}{5}, 0\right) = \left(\frac{1}{5}, 0\right)\).

6. **Use the property of the circle and diagonals:**
   - Since \(BD = AB\), and \(BD\) is a chord intersecting \(AC\) at \(P\), we can use the Power of a Point theorem.
   - The Power of a Point theorem states that \(AP \cdot PC = BP \cdot PD\).

7. **Calculate the length of \(CD\):**
   - Since \(P\) is on \(AC\), and \(AC\) is a diameter, \(P\) is the midpoint of \(BD\) because \(BD\) is perpendicular to \(AC\) at \(P\).
   - Let \(D = (x, y)\). Since \(D\) lies on the circle, \(x^2 + y^2 = 1\).
   - Since \(P\) is the midpoint of \(BD\), \(B\) and \(D\) are symmetric with respect to \(P\). Therefore, \(B = (2 \cdot \frac{1}{5} - x, -y) = \left(\frac{2}{5} - x, -y\right)\).

8. **Use the given length \(BD = AB\):**
   - Since \(BD = AB\), and \(AB\) is a chord of the circle, we can use the Pythagorean theorem in \(\triangle ABD\).
   - Let \(AB = BD = s\). Then \(s^2 = AB^2 = (x - (-1))^2 + y^2 = (x + 1)^2 + y^2\).

9. **Solve for \(CD\):**
   - Since \(CD\) is a chord of the circle, we can use the distance formula to find \(CD\).
   - \(CD = \sqrt{(x - 1)^2 + y^2}\).

10. **Substitute the values:**
    - Since \(x = \frac{1}{5}\) and \(y = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5}\),
    - \(CD = \sqrt{\left(\frac{1}{5} - 1\right)^2 + \left(\frac{2\sqrt{6}}{5}\right)^2} = \sqrt{\left(\frac{1}{5} - \frac{5}{5}\right)^2 + \left(\frac{2\sqrt{6}}{5}\right)^2} = \sqrt{\left(-\frac{4}{5}\right)^2 + \left(\frac{2\sqrt{6}}{5}\right)^2} = \sqrt{\frac{16}{25} + \frac{24}{25}} = \sqrt{\frac{40}{25}} = \sqrt{\frac{8}{5}} = \frac{2\sqrt{2}}{5}\).

The final answer is \(\boxed{\frac{2\sqrt{2}}{5}}\).