Equilibrium constant in terms of mole fractions
The initial mole fraction of $\ce{CO}$ present must be $y=0.5$.  Then the initial mole fraction of $\ce{H2O}$ present is also $y=0.5$, and the initial moles of both $\ce{H2}$ and $\ce{CO2}$ are 0. 

After equilibration, both $\ce{H2}$ and $\ce{CO2}$ are present in the amount $x$ and so by stoichiometry, $\ce{H2O}$ and $\ce{CO}$ must be present in the amount $y-x=0.5-x=\frac{1}{2}-x$.

Since the reaction doesn't change the number of moles of gas, it doesn't matter much if we use $K_p$ or $K_c$, so $K_p=\frac{x^2}{(\frac{1}{2}-x)^2}=\frac{x^2}{\frac{1}{4}-x+x^2}=~...$

I'm sure you can figure out the rest.