Calculate the potential difference V_a - V_b between points a and b...

The circuit labeled (b) is a single closed loop containing two capacitors and two batteries.

The equivalent capacitance $C_{eq}$ of the $2 \mu\text{F}$ and $4 \mu\text{F}$ capacitors in series is $C_{eq} = \dfrac{2 \times 4}{2 + 4} = \dfrac{4}{3} \mu\text{F}$.

Traversing the loop in the clockwise direction, the $12 \text{ V}$ battery acts to drive current clockwise, while the $24 \text{ V}$ battery opposes it. The net electromotive force is $\mathcal{E}_{net} = 24 - 12 = 12 \text{ V}$ in the counter-clockwise direction.

The steady-state charge $Q$ on the equivalent capacitor is $Q = C_{eq} \mathcal{E}_{net} = \left(\dfrac{4}{3}\right)(12) = 16 \mu\text{C}$.

This counter-clockwise flow of charge polarizes the capacitors such that the bottom plate of the $2 \mu\text{F}$ capacitor (at node $b$) is positive and the top plate (at node $a$) is negative.

The potential difference across the $2 \mu\text{F}$ capacitor is $V_c = \dfrac{Q}{C} = \dfrac{16}{2} = 8 \text{ V}$.

Since the plate at node $b$ is at a higher potential, $V_b - V_a = 8 \text{ V}$, which implies $V_a - V_b = -8 \text{ V}$.