First order gas phase reaction
$$A \to B + C$$
$$p_i$$ = initial pressure of gas A, $$p_t$$ = total pressure of the reaction mixture at time $$t$$
Expression of rate constant (k) is

Analyze the partial pressures of the components at different stages of the reaction:

$$Reaction:\ A_{\left(g\right)}+B_{\left(g\right)}\longrightarrow\ C_{\left(g\right)}$$

At t=0:              $$p_i$$          0          0

At time t:          $$p_i$$-x       x          x

Where $$x$$ is the decrease in the partial pressure of gas $$\text{A}$$ at time $$t$$.

Step 1: Relate Total Pressure $$(p_{t})$$ to $$x$$ The total pressure of the reaction mixture at time $$t$$ is the sum of the partial pressures of all gaseous species present in the container:

$$p_{t}=p_{\text{A}}+p_{\text{B}}+p_{\text{C}}$$

$$p_{t}=(p_{i}-x)+x+x$$ 

$$p_{t}=p_{i}+x$$

Solving for $$x$$:
$$x=p_{t}-p_{i}$$

Step 2: Find the Remaining Pressure of Reactant A $$(p_{\text{A}}$$)

Substitute the value of $$x$$ back into the expression for the pressure of $$\text{A}$$ at time $$t$$:
$$p_{\text{A}}=p_{i}-x$$

$$p_{\text{A}}=p_{i}-(p_{t}-p_{i})$$

$$p_{\text{A}}=2p_{i}-p_{t}$$

Step 3: Apply the First-Order Integrated Rate Law

For a first-order reaction, the integrated rate law in terms of partial pressures is:
$$k=\frac{2.303}{t}\log \frac{\text{Initial\ pressure\ of\ A}}{\text{Pressure\ of\ A\ at\ time\ }t}$$

Substituting the expressions we derived:
$$k=\frac{2.303}{t}\log \frac{p_{i}}{2p_{i}-p_{t}}$$

$$k=\frac{1}{t}\ln \frac{p_{i}}{2p_{i}-p_{t}}$$

Hence, Option A is correct