HINT: Use First order gas-phase reaction

Explanation:

Initially At time t  

 $\mathrm{A}\left(\mathrm{g}\right)$ $\to \mathrm{B}\left(\mathrm{g}\right)$ $+$ $\mathrm{C}\left(\mathrm{g}\right)$ $$
$$ ${\mathrm{P}}_{\mathrm{i}}$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $0$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $0$
${\mathrm{P}}_{\mathrm{i}}-\mathrm{x}$ $$ $$ $$ $$ $$ $\mathrm{x}$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $\mathrm{x}$
For the first order reaction 
\(\begin{aligned} & \mathrm{P}_{\mathrm{t}}=\mathrm{P}_{\mathrm{i}}-\mathrm{x}+\mathrm{x}=\mathrm{}\mathrm{P}_{\mathrm{i}}+\mathrm{x} \\ & \mathrm{x}=\mathrm{P}_{\mathrm{t}}-\mathrm{P}_{\mathrm{i}} \\ & \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_i}{\mathrm{P}_i-\mathrm{x}} \\ & =\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_i}{\mathrm{P}_i-\left(\mathrm{P}_{\mathrm{t}}-\mathrm{P}_i\right)} \\ & =\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{P}_i}{2 \mathrm{P}_i-\mathrm{P}_{\mathrm{t}}} \end{aligned}\)
According to the problem, the initial pressure of the system is P_i, and after a certain time t, the total pressure of the system increased by X units and became P_t. As X is partial pressure and we write our final expression  in initial  pressure and final pressure so most suitable answer will be the 2nd one not 1st as it uses the term X as the it is a partial pressure .

Hence, option second is the correct answer.