The reaction shown is:
$N_2+3H_2 = 2NH_3$
With a deduction of 4 moles of the $N_2/H_2$ mixture, 2 moles of Ammonia is obtained.
So, from 20% of the $N_2/H_2$ mixture, 10% Ammonia is produced, reducing the total mixture by 10%.
Given, the system is a closed vessel, Volume, V = constant and temperature, T = 100 atm.
$\therefore \quad \large p \propto n$
Since, the total mixture left is 90%, there will be a 10% decrease in the pressure on the mixture i.e. 90 atm.
Hence answer is (c)