When $NaN{O}_3$  is heated in a closed vessel, oxygen is liberated and $NaN{O}_2$  is left behind. At equilibrium

$2NaN{O}_3\left(s\right)\rightleftharpoons 2NaN{O}_2\left(s\right)+O_2\left(g\right)$

Appling law of Chemical Equilibrium. $K=\left[O_2\right(g\left)\right].$   Hence addition of $NaN{O}_2$  or $NaN{O}_3$  will not disturb the equilibrium $i.e.\left(a\right)$  and (b) are wrong. Only (c) and (d) are correct.