Examine the processes A and B displayed in the figure. It is...

The slope of a process on a $p-V$ diagram is given by $\dfrac{dp}{dV}$.

For an isothermal process, $pV = \text{constant}$, which gives $\dfrac{dp}{dV} = -\dfrac{p}{V}$.

For an adiabatic process, $pV^{\gamma} = \text{constant}$, which gives $\dfrac{dp}{dV} = -\gamma \dfrac{p}{V}$.

Since $\gamma > 1$, the adiabatic curve is steeper than the isothermal curve. For expansion from the same initial state, the adiabatic curve will lie below the isothermal curve.

In the given figure, curve B lies below curve A, meaning curve B is steeper than curve A.

Therefore, it is possible that process A is isothermal and process B is adiabatic.

It is also possible that both processes are adiabatic but for different gases. A gas with a higher value of $\gamma$ will have a steeper curve. For example, curve B could represent a monoatomic gas ($\gamma = 5/3$) and curve A could represent a diatomic gas ($\gamma = 7/5$).

If both were isothermal, they would follow the exact same curve $p = \dfrac{p_0 V_0}{V}$, which is not the case.