Graph of compressibility factor vs pressure when real gas is assigned Z=1
For a real gas, $(P+a/V^2)(V-b) = RT$

Now, since we are considering $Z=1$ for real gas, all the real gas equations will be reversed for the ideal gas. Meaning, we had to subtract $b$ for volume correction of real gas, for correction of ideal gas we need to add $b$. Same, with pressure. 

The equation then comes out to be: $(P-a/V^2)(V+b) = RT$

Now, for the graph, we can calculate the approximate values of $Z$ (relative to unity) for very low pressure, low pressure and high pressure.

Case 1: Very low pressure: $P \downarrow  \downarrow \implies V  \uparrow  \uparrow $

$(P-a/V^2) \approx P$ and $(V-b)\approx V \implies Z \approx 1$

Case 2: Low pressure $P \downarrow \implies V \uparrow$

$(V+b)\approx V$. On solving, we get $Z= 1+a/(RTV)\implies Z&gt;1$.

Case 3: Very high pressure $P \uparrow  \uparrow \implies V^2\downarrow  \downarrow$  

$(P - a/V^2) \approx P$. On solving, we get $Z= 1-Pb/(RT)\implies Z&lt;1$.



The graph would thus be (A), because it is the only one satisfying all these conditions.