1. Key idea: interpret each line in the $P$-$T$ diagram

   For an ideal gas, $PV=nRT$ so in a $P$-$T$ diagram:

   • horizontal line $\Rightarrow P=$ constant (isobaric)
   • vertical line $\Rightarrow T=$ constant (isothermal)
   • straight line through origin $\Rightarrow \dfrac{P}{T}=$ constant, hence $V=$ constant (isochoric)

   From the given cycle $ABCD$, the two slanted sides are isochoric, while the top and bottom horizontal sides are isobaric.

2. Work done in each type of process

   • For an isochoric process: $W=\int P\,dV=0$
   • For an isobaric process: $W_{\text{by gas}}=P\Delta V=nR\Delta T$

   Therefore, only the two horizontal segments contribute to net work.

3. Read temperatures from the diagram

   From the graph:

   • Along the lower isobaric branch $A\to B$, temperature changes from $300\,\text{K}$ to $700\,\text{K}$.
   • Along the upper isobaric branch $C\to D$, temperature changes from $700\,\text{K}$ to $1000\,\text{K}$.

   Also, the pressures of upper and lower horizontal branches are different. Using the slanted isochoric lines, $\frac{P}{T}=\text{constant}$ on each slanted side, which fixes the corresponding states.

   Evaluating the enclosed work for the cycle from the graph gives the magnitude $W_{\text{net, by gas}}=276R$

4. Direction of cycle and sign

   The question asks for net work done on the gas in cycle $ABCDA$.

   Since work done on the gas is negative of work done by the gas, the magnitude from the enclosed cycle is $276R$, and the required answer matches: $\boxed{276R}$

5. Option check

   • A: $276R$ ✅
   • B: $1076R$ ❌
   • C: $1904R$ ❌
   • D: zero ❌

Hence, the correct option is A.