Identify the nature of each process

From the statement:

• Along $A \to B$, temperature is constant at $T_1$.
• Along $C \to D$, temperature is constant at $T_2$, with $T_1 > T_2$.
• Processes $B \to C$ and $D \to A$ are adiabatic.

So the cycle is made of:

• two isothermal processes: $AB$ at $T_1$, $CD$ at $T_2$
• two adiabatic processes: $BC$ and $DA$

This is essentially a Carnot-type cycle.

Work done in the adiabatic processes

For an adiabatic process, since $Q=0$, from the first law: $\Delta U = -W$ where $W$ is work done by the gas.

For an ideal gas, $\Delta U = nC_V (T_f - T_i)$ Hence, $W = -\Delta U = nC_V (T_i - T_f)$

Calculate work in $B \to C$

In $B \to C$, temperature falls from $T_1$ to $T_2$. Therefore, $W_{BC} = nC_V (T_1 - T_2)$

Since $T_1 > T_2$, this is positive.

Calculate work in $D \to A$

In $D \to A$, temperature rises from $T_2$ to $T_1$. Therefore, $W_{DA} = nC_V (T_2 - T_1) = -nC_V (T_1 - T_2)$

Compare $W_{BC}$ and $W_{DA}$

From above, $W_{DA} = -W_{BC}$

So if the option uses $W_{AD}$ instead of $W_{DA}$, then reversing the direction changes the sign: $W_{AD} = -W_{DA} = W_{BC}$

Therefore, $W_{AD} = W_{BC}$

So Option B is correct.

Check the other options

Option A: $W_{AB} = W_{DC}$

Along isothermal process for ideal gas, $W = nRT \ln\left(\frac{V_f}{V_i}\right)$

For the Carnot cycle, $W_{AB} = nRT_1 \ln\left(\frac{V_B}{V_A}\right)$ and $W_{DC} = nRT_2 \ln\left(\frac{V_C}{V_D}\right)$

Using adiabatic connections, one gets $\frac{V_B}{V_A} = \frac{V_C}{V_D}$ hence the logarithmic factors are equal, but since $T_1 > T_2$, $W_{AB} > W_{DC}$ So A is false.

Option C: $W_{BC} + W_{DA} > 0$

But we found $W_{DA} = -W_{BC}$ Hence, $W_{BC} + W_{DA} = 0$ So C is false.

Option D: $W_{AB} < W_{CD}$

In the actual direction $C \to D$, this isothermal process is compression, so work done by gas is negative: $W_{CD} < 0$ while $W_{AB} > 0$. Therefore, $W_{AB} < W_{CD}$ is impossible. So D is false.

Final conclusion

The correct option is: $\boxed{B}$ i.e. $\boxed{W_{AD} = W_{BC}}$