For the equilibrium
$\mathrm{CO}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{CO}}_2+{\mathrm{H}}_2$
The relation between ${\mathrm{K}}_{\mathrm{p}}$ and ${\mathrm{K}}_{\mathrm{c}}$ at $25°\mathrm{C}$ and at $100°\mathrm{C}$ are

The effect of pressure depends on the $\Delta {\mathrm{n}}_{\mathrm{g}}$ of reaction
${\mathrm{\Delta n}}_{\mathrm{g}}={\mathrm{n}}_{\mathrm{P}\left(\mathrm{g}\right)}-{\mathrm{n}}_{{\mathrm{R}}_{\left(\mathrm{g}\right)}}$
${\mathrm{\Delta n}}_{\mathrm{g}}=0,$ No effoct of prossure.

${\mathrm{\Delta n}}_{\mathrm{g}}=+\mathrm{ve},$ P increases, the reaction takes place in a backward direction or vice-versa.

 ${\mathrm{\Delta n}}_{\mathrm{g}}=-\mathrm{ve},$P increases, the reaction takes place in a forward direction or vice-versa.

For the given reaction,

$\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{CO}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

${\mathrm{\Delta n}}_{\mathrm{g}}=0$

So, the equilibrium constant in terms of pressure becomes,

${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^{∆\mathrm{n}}; \mathrm{at} 25°\mathrm{C}$

And

${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}} \mathrm{at} 100°\mathrm{C}$

Thus, option 4 is correct.