To solve the problem, we need to find the value of the expression $\sin A \sin B \left(\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} + \cot \frac{D}{2}\right)^2$ for the given cyclic quadrilateral $ABCD$ with side lengths $AB = 2$, $BC = 3$, $CD = 5$, and $AD = 4$.

1. **Calculate the semi-perimeter $s$ of the quadrilateral:**
$$s = \frac{AB + BC + CD + DA}{2} = \frac{2 + 3 + 5 + 4}{2} = 7$$

2. **Use Brahmagupta's formula to find the area $K$ of the cyclic quadrilateral:**
$$K = \sqrt{(s - AB)(s - BC)(s - CD)(s - DA)} = \sqrt{(7 - 2)(7 - 3)(7 - 5)(7 - 4)} = \sqrt{5 \cdot 4 \cdot 2 \cdot 3} = \sqrt{120} = 2\sqrt{30}$$

3. **Find the radius $R$ of the circumcircle using the formula $K = \frac{1}{2} \cdot d_1 \cdot d_2 \cdot \sin \theta$, where $d_1$ and $d_2$ are the diagonals and $\theta$ is the angle between them. However, we can use another approach to find $R$ directly:**
$$R = \frac{abc}{4K} \quad \text{(for a triangle, but we need to adapt for quadrilateral)}$$

4. Use the known identity for cyclic quadrilaterals:
$$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} + \cot \frac{D}{2} = \frac{4s}{r}$$
where $r$ is the radius of the incircle of the quadrilateral. For a cyclic quadrilateral, the radius $r$ can be found using:
$$r = \frac{K}{s} = \frac{2\sqrt{30}}{7}$$

5. Substitute the values into the identity:
$$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} + \cot \frac{D}{2} = \frac{4 \cdot 7}{\frac{2\sqrt{30}}{7}} = \frac{28 \cdot 7}{2\sqrt{30}} = \frac{196}{2\sqrt{30}} = \frac{98}{\sqrt{30}} = \frac{98\sqrt{30}}{30}$$

6. Square the sum of cotangents:
$$\left(\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} + \cot \frac{D}{2}\right)^2 = \left(\frac{98\sqrt{30}}{30}\right)^2 = \frac{9604 \cdot 30}{900} = \frac{288120}{900} = 320.1333$$

7. **Calculate $\sin A \sin B$:**
Since $ABCD$ is cyclic, we use the fact that $\sin A \sin B = \sin C \sin D$. However, without specific angles, we assume symmetry and use the product of sines for the quadrilateral:
$$\sin A \sin B = \sin C \sin D = \frac{K}{R^2}$$

8. Combine all parts to find the final expression:
$$\sin A \sin B \left(\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} + \cot \frac{D}{2}\right)^2 = \frac{2\sqrt{30}}{R^2} \cdot 320.1333$$

The final answer is $\boxed{a + b}$.