Answer
$$ N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$ $K_p = 1.3 \times 10^{-29}$$
Work Step by Step
$NO(g) + \frac 1 2 Br_2(g) \rightleftharpoons NOBr(g) $ $K_p = 5.3$
$2\space NO(g) \rightleftharpoons N_2(g) + O_2(g)$ $K_p = 2.1 \times 10^{30}$
1. In the final equation, the $N_2$ molecule is on the reactants side. Thus, we should invert the second equation. The new $K_p$ is equal to the multiplicative inverse of the old one.
$K_p = \frac 1 {2.1 \times 10^{30}} = 4.8 \times 10^{-31}$
$NO(g) + \frac 1 2 Br_2(g) \rightleftharpoons NOBr(g) $ $K_p = 5.3$
$N_2(g) + O_2(g) \rightleftharpoons 2\space NO(g)$ $K_p = 4.8 \times 10^{-31}$
2. In the final equation the coefficient that goes with $Br_2$ is $1$, so we should multiply all the first equation by 2. The new $K_p$ is equal to $5.3$ to the power o 2.
$K_p = 5.3^2 = 28$
$2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g) $ $K_p = 28$
$N_2(g) + O_2(g) \rightleftharpoons 2\space NO(g)$ $K_p = 4.8 \times 10^{-31}$
3. Now, we can add the equations. The final $K_p$ is equal to the multiplication of $28$ and $4.8 \times 10^{-31}$
$K_p = 28 \times 4.8 \times 10^{-31} = 1.3 \times 10^{-29}$
$2NO(g) + N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2\space NO(g) + 2NOBr(g)$ $K_p = 1.3 \times 10^{-29}$
$$ N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2NOBr(g)\space K_p = 1.3 \times 10^{-29}$$
