For the reaction : ${\mathrm{H}}_{2\left(\mathrm{g}\right)} + {\mathrm{CO}}_{2\left(\mathrm{g}\right)} \leftrightharpoons {\mathrm{CO}}_{\left(\mathrm{g}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)},$ if the initial concentration of 

$\left[{\mathrm{H}}_2\right] = \left[{\mathrm{CO}}_2\right]$ and x moles / litre of hydrogen is consumed at equilibrium, the correct expression of ${\mathrm{K}}_{\mathrm{P}}$ is

                              ${\mathrm{H}}_{2\left(\mathrm{g}\right)} + {\mathrm{CO}}_{2\left(\mathrm{g}\right)} \leftrightharpoons {\mathrm{CO}}_{\left(\mathrm{g}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

Initial conc.               1            1                    0              0

At equilibrium        (1 - x)     (1 - x)               x               x

${\mathrm{K}}_{\mathrm{P}}=\frac{{\mathrm{p}}_{\mathrm{CO}} . {\mathrm{p}}_{{\mathrm{H}}_2\mathrm{O}}}{{\mathrm{p}}_{{\mathrm{H}}_2.}{\mathrm{p}}_{{\mathrm{CO}}_2}}$ = $\frac{\mathrm{x}.\mathrm{x}}{\left(1-\mathrm{x}\right) \left(1-\mathrm{x}\right)}= \frac{{\mathrm{x}}^2}{{\left(1-\mathrm{x}\right)}^2}$