B

To find the equivalent capacitance between terminals 1 and 4 in the given configuration, we begin by analyzing the arrangement of capacitors.

1. **Identify the Connections:** There are three branches connected in parallel between terminals 1 and 4. Each branch consists of two capacitors in series.

2. **Calculate Capacitance of Series Capacitors:** For two capacitors in series, the equivalent capacitance (C_s) is given by:
$$ C_s = \frac{C \cdot C}{C + C} = \frac{C}{2} $$
where C = 15 µF (given).
So,
$$ C_s = \frac{15 \mu F}{2} = 7.5 \mu F $$
Each branch has a capacitance of 7.5 µF.

3. **Calculate Equivalent Capacitance of Parallel Capacitors:** Since there are three such branches in parallel, the total equivalent capacitance (C_eq) is given by:
$$ C_{eq} = C_s + C_s + C_s = 3 C_s = 3 \times 7.5 \mu F = 22.5 \mu F $$

4. **Final Calculation:** Therefore, the equivalent capacitance between terminals 1 and 4 is 22.5 µF.

Thus, the equivalent capacitance is **B) 22.5 µF**.