$\begin{aligned}
& \text { (b) } \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\
& \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) \\
& \Rightarrow \Delta \mathrm{n}_{\mathrm{g}}=(1+1)-(1+1)=0 \\
& \Rightarrow \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} .
\end{aligned}$
Now, $40 \%$ of mass of water reacted with CO.
$\Rightarrow \alpha=0.4$
\begin{array}{|c|c|c|c|c|}
\hline & \mathrm{H}_2 \mathrm{O} & \mathrm{CO} & \mathrm{H}_2 & \mathrm{CO}_2 \\
\hline Initial moles & 1.0 & 1.0 & 0.0 & 0.0 \\
\hline At equilibrium & \begin{array}{c}
1-0.4 \\
=0.6
\end{array} & \begin{array}{c}
1-0.4 \\
=0.6
\end{array} & 0.4 & 0.4 \\
\hline
\end{array}
Thus, $\left[\mathrm{H}_2 \mathrm{O}\right]=\frac{0.6}{1.0}=0.6 \mathrm{M},[\mathrm{CO}]=0.6 \mathrm{M}$,
$\begin{aligned}
& {\left[\mathrm{H}_2\right]=0.4 \mathrm{M},\left[\mathrm{CO}_2\right]=0.4 \mathrm{M}} \\
& \Rightarrow \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{H}_2\right]\left[\mathrm{CO}_2\right]}{\left[\mathrm{H}_2 \mathrm{O}\right][\mathrm{CO}]}=\frac{(0.4)(0.4)}{(0.6)(0.6)}=0.444
\end{aligned}$
Thus, $K_p=0.444$.