By observing the data given, by doubling the pressure, half-life is reduced to half. This result is following second order kinetics. Mathematically it can be proved as follows,

The half-life and initial pressure of ${\mathrm{n}}^{\mathrm{th}}$ order reaction can be related as follows,

${\mathrm{t}}_{\frac{1}{2}}\propto {\left({\mathrm{P}}_{\mathrm{o}}\right)}^{1-\mathrm{n}}$

$\frac{{\left({\mathrm{t}}_{{\displaystyle \frac{1}{2}}}\right)}_1}{{\left({\mathrm{t}}_{\frac{1}{2}}\right)}_2}=\frac{{\left({\mathrm{P}}_0\right)}_1^{1-\mathrm{n}}}{{\left({\mathrm{P}}_0\right)}_2^{1-\mathrm{n}}}$

$\Rightarrow \left(\frac{4}{2}\right)={\left(\frac{50}{100}\right)}^{1-\mathrm{n}}$

$\Rightarrow 2={\left(\frac{1}{2}\right)}^{1-\mathrm{n}}$

$\Rightarrow 2={\left(2\right)}^{\mathrm{n}-1}$

$\Rightarrow \mathrm{n}-1=1$

$\Rightarrow \mathrm{n}=2$

Hence, the order of the reaction is second order.