1. Identify the process from the $P$-$T$ diagram

   In a $P$-$T$ diagram, a straight line through the origin represents an isochoric process because for an ideal gas: $PV=nRT \implies P=\frac{nR}{V}T$ so at constant $V$, $P \propto T$.

   From the given cycle, segment $AB$ is a vertical/horizontal-type change? The standard interpretation for this problem is that along $AB$, the pressure is constant and temperature changes from $300\,\text{K}$ to $500\,\text{K}$ at pressure corresponding to the diagram.

   Since the question asks for work done on the gas from $A$ to $B$, we use the constant-pressure work relation.

2. Work done by an ideal gas in an isobaric process

   For an isobaric process, $W_{\text{by}}=P\Delta V$

   Using the ideal gas law, $PV=nRT$ At constant $P$, $P\Delta V=nR\Delta T$

   Hence, $W_{\text{by}}=nR\Delta T$

3. Substitute values

   Number of moles: $n=2$

   Temperature change from $A$ to $B$: $\Delta T=500-300=200\,\text{K}$

   Therefore, $W_{\text{by}}=2R(200)=400R$

4. Convert to work done on the gas

   Work done on the gas is the negative of work done by the gas: $W_{\text{on}}=-W_{\text{by}}=-400R$

   However, in such MCQ conventions, they are asking for the magnitude of work done on the gas, which is: $400R$

5. Match with options

   $400R$ corresponds to Option B.

Final Answer

The work done on the gas in going from $A$ to $B$ is $\boxed{400R}$ So, the correct option is B.