Let’s first draw the diagram for a better understanding.
Let the distance between $\text{P},$ and $\text{Q}$ be $\text{‘D’}$ meter.
Given that, $\text{S}_{\text{P}}:\text{S}_{\text{Q}} = 3:4 \Rightarrow {\color{Blue}{\boxed{\frac{\text{S}_{\text{P}}}{\text{S}_{\text{Q}}} = \frac{3}{4}}}}$
We know that, ${\color{Green}{\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \Rightarrow \text{Speed} \propto \text{Distance (Time constant)}}}$
$\implies {\color{Purple}{\boxed{\frac{\text{S}_{1}}{\text{S}_{2}} = \frac{\text{D}_{1}}{\text{D}_{2}}}}}$
Now, $\dfrac{3}{4} = \dfrac{500-140}{500-\text{D}}$
$\Rightarrow \dfrac{3}{4} = \dfrac{360}{500-\text{D}}$
$\Rightarrow 3(500-\text{D}) = 4 \times 360$
$\Rightarrow 1500-3\text{D} = 1440$
$\Rightarrow 3\text{D} = 60$
$\Rightarrow {\color {Blue}{\boxed{\text{D = 20}}}}$
$\therefore$ The distance between $\text{P}$ and $\text{Q},$ when $\text{P}$ wins the race is $20\;\text{meters}.$
Correct Answer $:\text{A}$