Explanation

(b)
$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$
Moles of fraction of $NH_{3}\left( X_{NH_{3}} \right) = 0.6$
Mole fraction of $N_{2}\ and\ H_{2} = 1 - 0.6 = 0.4$
Total no. of mole of $N_{2}\ and\ H_{2} = 1 + 3 = 4$
Then,
$X_{N_{2}} = \frac{1}{4} \times 0.4 = 0.1$
$X_{H_{2}} = \frac{3}{4} \times 0.4 = 0.3$
Partial pressure of $N_{2}\left( p_{N_{2}} \right)$
$\ \ \ \ \ \ = 0.1 \times 10 = 1\ atm$
Partial pressure of $H_{2}\left( p_{H_{2}} \right)$
$\ \ \ \ \ = 0.3 \times 10 = 3\ atm$
Partial pressure of $NH_{3}\left( p_{NH_{3}} \right)$
$\ \ \ \ \ = 0.6 \times 10 = 6\ atm$
Dissociation of ammonia
$2NH_{3} \rightleftharpoons N_{2} + 3H_{2}$
$K_{p} = \frac{p_{N_{2}} \times p_{H_{2}}^{3}}{p_{NH_{3}}^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1 \times (3)^{3}}{(6)^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{27}{36} = 0.75\ atm^{2}$