1. For a reaction of order $n$, the half-life depends on the initial concentration (or initial pressure for a gas) as:

$t_{1/2} \propto \frac{1}{a^{\,n-1}}$

where $a$ is the initial concentration/pressure.

Since for a gas at constant temperature,

$a \propto p$

we can write:

$t_{1/2} \propto p^{-(n-1)}$

2. From the given data:

• When $p$ doubles from $50$ to $100$, relative $t_{1/2}$ changes from $4$ to $2$.
• When $p$ doubles again from $100$ to $200$, $t_{1/2}$ changes from $2$ to $1$.
• When $p$ doubles again from $200$ to $400$, $t_{1/2}$ changes from $1$ to $0.5$.

So, every time pressure doubles, half-life becomes half.

Thus,

$t_{1/2} \propto \frac{1}{p}$

3. Compare with the general relation:

$t_{1/2} \propto p^{-(n-1)}$

Given:

$p^{-(n-1)} = p^{-1}$

Therefore,

$n-1=1$ $n=2$

4. Hence the reaction is of second order.

So the correct option is:

$\boxed{\text{A: }2}$