In the reaction, P + Q → R + S the time taken for 75 % reaction of P is twice the time taken for 50 %…

The reaction is:

$\text{P}+\text{Q}\to \text{R}+\text{S}$
$\mathrm{rate} \mathrm{of} \mathrm{the} \mathrm{reaction}=\mathrm{k}{\left[\mathrm{P}\right]}^{\mathrm{x}}{\left[\mathrm{Q}\right]}^{\mathrm{y}}$
The concentration of Q varies with time as per the graph by the equation.
$\left[\text{Q}\right]=\left[{\text{Q}}_0\right]-\text{kt }$ i.e. Zero-order kinetics.

$\Rightarrow \mathrm{y}=0$
With respect to reactant $\mathrm{P}$

let time taken for $50\%$ completion of the reaction is $\mathrm{a}$.

Then the time is taken for $75\%$  complication will be $2\mathrm{a}.$

$\mathrm{P}\stackrel{{\mathrm{t}}_{50\%}}{\to } \frac{\mathrm{P}}{2}\stackrel{{\mathrm{t}}_{75\%}}{\to } \frac{\mathrm{P}}{4}$

at

$$\begin{gathered} \mathrm{t}=0 ; [\mathrm{P}{]}_{0\%}= \mathrm{P} \\ \mathrm{t}= {\mathrm{t}}_{50\% }; [\mathrm{P}{]}_{50\%}= \frac{\mathrm{P}}{2} \\ \mathrm{t}= {\mathrm{t}}_{75\% }; [\mathrm{P}{]}_{75\%}= \frac{\mathrm{P}}{4} \end{gathered}$$
As we can see, in each time interval of 'a' min the reactant concentration is changing by factor half. therefore rate constant is independent of the concentration of the reactant $\mathrm{P}$. hence with respect of reactant $\mathrm{P}$, order of the reaction is $1$. 

$\Rightarrow \mathrm{x}=1$
Rate law expression is $\mathrm{rate}=\mathrm{k}{\left[\mathrm{P}\right]}^1{\left[\mathrm{Q}\right]}^0$
The overall order of reaction $=x+y = 1+0= 1$