A fluid undergoes a thermodynamic process such that its pressure P...

The bulk modulus is defined as $B = -V \dfrac{dP}{dV}$.

Given the parametric equations:
$P = a t^3$
$V = \dfrac{b}{t^2}$

Differentiate both with respect to $t$:
$\dfrac{dP}{dt} = 3at^2$
$\dfrac{dV}{dt} = -\dfrac{2b}{t^3}$

Now, find $\dfrac{dP}{dV}$ using the chain rule:
$\dfrac{dP}{dV} = \dfrac{\dfrac{dP}{dt}}{\dfrac{dV}{dt}} = \dfrac{3at^2}{-\dfrac{2b}{t^3}} = -\dfrac{3a}{2b}t^5$

Substitute this into the bulk modulus formula:
$B = -V \dfrac{dP}{dV} = -\left(\dfrac{b}{t^2}\right) \left(-\dfrac{3a}{2b}t^5\right)$
$B = \dfrac{3a}{2}t^3$

Since $P = a t^3$, we can substitute $P$ back into the expression:
$B = \dfrac{3}{2}P$

Answer: $\dfrac{3P}{2}$