Ammonia decomposition on platinum surface
Thanks to Poutnik's comment, I arrived to the answer and I would like to share it.

$$\ce{2NH_3 -&gt;N_2 + 3H_2}$$
$$\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{N_2}=\frac{K_{N_2}p(N_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{H_2}=\frac{K_{H_2}p(H_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$

Since the hydrogen binds strongly on the surface, we can safely determine that $K_{H_2}&gt;&gt;1$, wich means the denominator $1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)\approx K_{H_2}p(H_2)$. So the surface coverage of ammonia can be written as $\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$. By definition $v=-\frac{dp(NH_3)}{dt}=k_0\theta_{NH_3}=k_0\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$, defining $k=k_0\frac{K_{NH_3}}{K_{H_2}}$ we arrive at the desired equation.

Feel free to edit.