Show that decomposition of AsH3 follows first order kinetics
In a first order reaction $\ce{A \rightarrow B + C}$ with rate constant k the rate equation is 
$$ {\frac{d\ce{[A]}}{dt} = - k\ce{[A]} } $$
which  if integrated to give the concentration vs time where the initial concentration is $\ce{[A_0]}$ produces
 $$\ln\left(\frac{\ce{[A]}}{\ce{[A_0]}}\right)=-kt$$ or $$ \ce{[A]= [A_0]}e^{-kt}$$

Thus a log of the ratio concentrations or pressures (or any measurement proportional to concentration) vs time should be a straight line if the reaction is first order.

In your case you are observing the product which appears as the $\ce{AsH3}$ disappears therefore
$$ {\frac{d\ce{[B]}}{dt} = + k\ce{[A]} } $$

You can substitute for $\ce{[A]}$ and integrate again to get $\ce{[B]}$.

$$\ce{[B]-[B_0]=[A_0]}(1-e^{-kt})$$
However the total pressure is measured which is the sum of $\ce{2[A] + 3[B]}$ and if $\ce{[B_0] =0}$ then 

$$\ce{3[B] + 2[A] =3[A_0]}(1-e^{-kt}) + \ce{2[A_0]}e^{-kt}$$
if the measured pressure is proportional to sum of concentrations with $p_0$ as initial pressure then 

$$p = p_0(3-e^{-kt})$$

At $t=0$ the initial pressure is $2p_0$ which is the value given in the table. At very long times the pressure becomes constant at $3p_0$ and calculating $3p_0 - p$, rearranging and taking logs produces

$$\ln\left(\frac{3p_0 - p}{p_0}\right) =  -kt $$

which when plotted is the 'normal' equation for a first order equation.