b
Mean force path, $\lambda=\frac{11}{P} \sqrt{\frac{\pi R T}{2 M}}$
where, $\mu=$ viscerity $P=$ Pressure $R=$ gass constant $T=$ Temperature $M=$Molecular weight
$\Rightarrow \lambda \alpha \frac{1}{p}$
If $p \propto \frac{1}{\lambda}$
If we double the mean force
Path, the pressure will be $1 / 2$.
So, The correct answer is $\frac{p}{2}$