The initial rate of a chemical reaction, P + Q arrow Products , was...

Let the rate law be $\text{Rate} = k[P]^x[Q]^y$.

From Experiment 1 and 2, when $[Q]$ is kept constant at $0.02 \text{ mol L}^{-1}$, the concentration of $[P]$ is increased from $0.02$ to $0.06$ (by a factor of $3$). The rate increases from $4.0 \times 10^{-5}$ to $1.2 \times 10^{-4}$ (by a factor of $3$).
$\dfrac{\text{Rate}_2}{\text{Rate}_1} = \dfrac{k(0.06)^x(0.02)^y}{k(0.02)^x(0.02)^y}$
$\dfrac{1.2 \times 10^{-4}}{4.0 \times 10^{-5}} = \left(\dfrac{0.06}{0.02}\right)^x$
$3 = 3^x \Rightarrow x = 1$

From Experiment 2 and 3, when $[P]$ is kept constant at $0.06 \text{ mol L}^{-1}$, the concentration of $[Q]$ is increased from $0.02$ to $0.04$ (by a factor of $2$). The rate increases from $1.2 \times 10^{-4}$ to $4.8 \times 10^{-4}$ (by a factor of $4$).
$\dfrac{\text{Rate}_3}{\text{Rate}_2} = \dfrac{k(0.06)^x(0.04)^y}{k(0.06)^x(0.02)^y}$
$\dfrac{4.8 \times 10^{-4}}{1.2 \times 10^{-4}} = \left(\dfrac{0.04}{0.02}\right)^y$
$4 = 2^y \Rightarrow y = 2$

Therefore, the rate law is $\text{Rate} = k[P][Q]^2$.

Answer: $\text{Rate} = k[P][Q]^2$