$N_2\iff 2N$
Initial Moles of $N_2 = \large\frac{1.4}{28}$
Initial Mole of 2N = 0
Mole after dissociation of $N_2 = \large\frac{1.4}{28}\times \large\frac{70}{100}$
Moles after dissociation of 2N = $\large\frac{1.4}{28}\times \large\frac{2\times30}{100}$
Total Mole = $\large\frac{1.4}{28}\times\large\frac{70}{100}+\large\frac{1.4\times60}{100\times28}$
$=\large\frac{1.4}{28}\times[\large\frac{130}{100}]$
$P\times5 = \large\frac{1.4\times130}{28\times100}\times0.0821\times1800$
P = 1.92 atm
Hence answer is (b)