When a body moves under the influence of a constant power, the relationship between displacement and time can be established through the concept of power. Power (P) is defined as the rate at which work is done, and it can also be expressed in terms of force (F) and velocity (v) as $ P = F \cdot v $.

For a constant power P and assuming the force acts in the direction of the velocity, we can analyze how displacement (s) changes with time (t). Since force can also be written as $ F = \frac{d(mv)}{dt} $ for a constant mass m, this simplifies to $ F = m \frac{dv}{dt} $, because mass doesn't change with time for most cases. Integrating force over a distance gives work (W), and power is the rate of doing work, thus we can connect these concepts.

The kinetic energy (K.E) of the body is given by $ K.E = \frac{1}{2}mv^2 $, and the work done by the force is equal to the change in kinetic energy. Considering power is constant, $ P = \frac{dW}{dt} = \frac{d(\frac{1}{2}mv^2)}{dt} $. Rearranging terms to focus on velocity and integrating with respect to time will give us a relation involving velocity and time.

For a constant mass system, and using $ P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v $, and knowing that $ P = \text{constant} $, we rearrange to find the relationship between velocity and time.

Given $ P = m \cdot v \cdot \frac{dv}{dt} $, we rearrange to $ \frac{P}{m} dt = v dv $. Integrating both sides where the initial condition is when $ t = 0, v = 0 $, we get $ \frac{P}{m} t = \frac{1}{2} v^2 $, solving for $ v $ gives $ v \propto t^{1/2} $, so $ v = k \cdot t^{1/2} $ for some constant $ k $.

The displacement $ s $ is obtained by integrating the velocity with respect to time, $ s = \int v dt = \int k \cdot t^{1/2} dt = \frac{2}{3}k \cdot t^{3/2} $. Therefore, the displacement $ s $ is proportional to $ t^{3/2} $.

The correct answer is Option B, $ t^{3/2} $.