For the gaseous phase reaction $2A+B\rightleftharpoons 2C+D$ , initially there are 2 mole each of A & B. If 0.4 mol of D is present at equilibrium at a given T & P, incorrect relationship is

$$\begin{gathered} 2\mathrm{A} + \mathrm{B} \rightleftharpoons 2\mathrm{C} + \mathrm{D} \\ 2 2 0 0 \mathrm{Initial} \\ (2-2\mathrm{x}) (2-\mathrm{x}) 2\mathrm{x} \mathrm{x} \mathrm{Equilibrium} \end{gathered}$$

But D at Equilibrium = 0.4

So x = 0.4

$$\begin{gathered} A=2-2(0.4)=1.2 \\ B=2-x=2-0.4=1.6 \\ C=2(0.4)=0.8 \\ D=0.4 \\ {P}_B>P_A>P_C>P_C \\ {P}_C=2P_D \\ {P}_A=3\frac{P_B}{4} \end{gathered}$$