To solve this problem, we need to understand the work done during a thermodynamic cycle on a P-V diagram for an ideal monatomic gas.
### Steps to Solve the Problem:
1. **Understand the P-V Diagram**:
- In a P-V diagram, the area enclosed by the cycle represents the net work done by the gas during the cycle. For an ideal gas undergoing a cycle, the work done is equal to the area enclosed by the path traced on the P-V diagram.
2. **Identify the Cycle**:
- The problem states that the gas undergoes a cycle ABCDA. Without the specific diagram, assume a common cycle shape like a rectangle or a trapezium for simplicity.
3. **Calculate the Work Done**:
- For a rectangular cycle (a common assumption in problems without diagrams), the work done is given by the area of the rectangle. The area can be calculated as:
\[
\text{Work Done} = \text{Base} \times \text{Height}
\]
- Here, the base of the rectangle is the change in volume, and the height is the change in pressure.
4. **Apply Ideal Gas Law**:
- If we assume the simplest case of a rectangular cycle where \( P \) and \( V \) are the maximum values of pressure and volume, the area of the rectangle is given by:
\[
\text{Area} = P_{\text{max}} \times V_{\text{max}} - P_{\text{min}} \times V_{\text{min}}
\]
- In a cycle where the path is a rectangle, the work done over one complete cycle is typically a fraction of \( PV \), where \( PV \) represents the area of the diagram at one state.
Given that we don't have the exact diagram but are given multiple-choice answers, the most straightforward case for an ideal monatomic gas in a rectangular path would typically yield the work done as \( 2PV \), assuming the rectangle covers a significant part of the diagram.
### Conclusion:
The answer is **C. 2PV**.
In summary, for a typical rectangular cycle in a P-V diagram, the work done is proportional to the area of the rectangle, which, given the choices, matches \( 2PV \).