(a)
For the reaction, we know that
$K_{p} = K_{c}(RT)^{\mathrm{\Delta}n_{g\ }}$
where, $\mathrm{\Delta}n_{g} =$ Number of gaseous moles of product – Number of gaseous moles of reactant
Hence, if $K_{p} > K_{c}$, then $\mathrm{\Delta}n$ must be positive.
It means that there is a decrease in volume of reaction, hence low pressure will favour the forward reaction.