In the circuit shown, the potential difference between A and B is

Since terminals A and B are open-circuited, the net current entering node D from A and leaving node C to B is zero:

$$I_{AD} = 0 \implies V_A - V_D = 0 \implies V_A = V_D$$

$$I_{CB} = 0 \implies V_C - V_B = 0 \implies V_C = V_B$$

The three branches between nodes D and C are connected in parallel.

$$\varepsilon_1 = 1\text{ V}, \quad r_1 = 1\ \Omega$$

$$\varepsilon_2 = 2\text{ V}, \quad r_2 = 1\ \Omega$$

$$\varepsilon_3 = 3\text{ V}, \quad r_3 = 1\ \Omega$$

Equivalent EMF ($$\varepsilon_{\text{eq}}$$) across nodes D and C:

$$\varepsilon_{\text{eq}} = \frac{\sum \frac{\varepsilon_i}{r_i}}{\sum \frac{1}{r_i}}$$

$$\varepsilon_{\text{eq}} = \frac{\frac{1}{1} + \frac{2}{1} + \frac{3}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}} = \frac{1 + 2 + 3}{3} = \frac{6}{3} = 2\text{ V}$$

$$V_A - V_B = V_D - V_C = \varepsilon_{\text{eq}} = 2\text{ V}$$