K_C for $N_2 + O_2 \leftrightharpoons 2NO$ at certain temperature is 1.6 × 10^-3, then K_P for $NO \leftrightharpoons 1/2N_2 + 1/2O_2$ at same temperature will be.

Given: it is given that $K_c=1.6\times {10}^{-3 }atm$ for reaction 1st: $N_2+O_2\rightleftharpoons 2NO$

We have to find $K_p$ for reaction 2nd: $NO\rightleftharpoons 1/2N_2+1/2O_2$

Formula Used: $K_p=K_c(RT{)}^{\Delta n};\Delta n=n_p-n_r$ 

Explanation: For reaction 1st, $n_p=2$ and $n_r=1+1=2$ ; therefore, $\Delta n=2-2=0$ and $K_p=K_c=1.6\times {10}^{-3}$

In reaction 2nd, NO gives $1/2N_2+1/2O_2$ i.e., ½ of the inverse of reaction 1st,

Therefore, $K_p\text{'}=1/(K_p{)}^{1/2}=1/(1.6\times {10}^{-3}{)}^{1/2}=25$

Hence, the correct option is A i.e., 25.