B

### Part (a) Calculation:
1. **Identify the configuration**: The circuit consists of capacitors connected in series and parallel.
2. **Calculate equivalent capacitance**:

• The two 2 µF capacitors are in parallel, giving a total capacitance of:
  $$ C_p = C_1 + C_2 = 2 ext{ µF} + 2 ext{ µF} = 4 ext{ µF} $$
• This total (4 µF) is in series with the 4 µF capacitor, leading to an equivalent capacitance of:
  $$ \frac{1}{C_{eq1}} = \frac{1}{4 ext{ µF}} + \frac{1}{4 ext{ µF}} $$
  $$ C_{eq1} = 2 ext{ µF} $$
• Next, we combine this with the 12V source. The total voltage across this equivalent capacitance is the same as the source voltage, i.e., 12V.
• The charge on each capacitor can be found using $Q = C \times V$. For the 2 µF capacitors, we find:
  $$ Q = 2 \text{ µF} \times 12V = 24 \text{ µC} $$
• The voltage across the capacitors can then be calculated as:
  $$ V = \frac{Q}{C} = \frac{24 \text{ µC}}{2 \text{ µF}} = 12V $$

3. **Voltage at point a and b**: Since both capacitors have the same voltage across them due to their parallel connection, the potential difference is:
$$ V_a - V_b = 0V $$

### Part (b) Calculation:
1. **Identify the configuration**: The circuit is a series connection of capacitors.
2. **Calculate equivalent capacitance**: The total capacitance is:
$$ C_{eq} = \frac{1}{\frac{1}{2 \text{ µF}} + \frac{1}{4 \text{ µF}}} = 1.33 \text{ µF} $$
3. **Total voltage across the circuit**: The total voltage provided is 24V.
4. **Voltage across each capacitor**: The voltage division rule gives:
$$ V = QC $$
5. Calculate charge for equivalent capacitance:
$$ Q = C_{eq} \times V_{total} = 1.33 \text{ µF} \times 24V = 32 \text{ µC} $$
6. Voltage across the capacitor a and b:

• For capacitor at a:
  $$ V_a = \frac{Q}{C}=\frac{32 \text{ µC}}{2 \text{ µF}} = 16V $$
• For capacitor at b:
  $$ V_b = \frac{Q}{C}=\frac{32 \text{ µC}}{4 \text{ µF}} = 8V $$

7. **Potential difference**:
$$ V_a - V_b = 16V - 8V = 8V $$
Thus, the potential difference V_a - V_b is 8V.
Therefore, for part (b), the correct option is **B**.