Given that, the speed of the boat in still water is $S_{b} = 12\;\text{km/hr}$ and the speed of the current is $S_{c} = 4\;\text{km/hr}.$
In upstream : The speed $S_{1} = S_{b} – S_{c} = 12-4 = 8\;\text{km/hr}$
In downstream : The speed $S_{2} = S_{b} + S_{c} = 12+4 = 16\;\text{km/hr}$
And, the total time $T = 4\;\text{hrs}$
We know that, $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$
Let’s distance from $P$ to $Q$ is $D.$
Now, $\dfrac{D}{S_{1}} + \dfrac{D}{S_{2}} = T$
$\implies \dfrac{D}{8} + \dfrac{D}{16} = 4$
$\implies \dfrac{2D+D}{16} = 4$
$\implies 3D = 64$
$\implies D = \dfrac{64}{3}\;\text{km} = 21\dfrac{1}{3}\;\text{km}.$
So, the correct answer is $(C).$