$\Delta H^{\circ}$ for the given reaction is $-46\: kJ$
$\Delta S^{\circ} = 2 \times S^{\circ}(NH_3) - 3 \times S^{\circ}(H_2) – S^{\circ}(N_2)$
$ = 2(192) – 3(130) – 191 = -197\: J/K$
$ \Delta G^{\circ} = \Delta H^{\circ} – T \Delta S^{\circ} = -46 – \bigg(\large\frac{-197}{1000} \bigg) $$(298) = -12.706\: kJ$
Now, $ \Delta G^{\circ} = -2.303\: RT \: logK_P$
$ \Rightarrow log\: K_P =\large\frac{ \Delta G^{\circ}}{-2.303RT }$$= \large\frac{(-12.706 \times 1000)}{(-2.303 \times 8.314 \times 298)}$$ = 2.226$
Ans : (A)