For an adiabatic process, the following relation holds:

$$P V^{\gamma} = \text{constant}$$

where P is the pressure, V is the volume, and $$\gamma = \frac{C_p}{C_v}$$.

We are given that the pressure is proportional to the cube of the absolute temperature:

$$P \propto T^3$$.

Using the ideal gas law, $$PV = nRT$$, we can rewrite this as:

$$V \propto \frac{T}{P} \propto \frac{T}{T^3} \propto \frac{1}{T^2}$$.

Substituting this into the adiabatic relation, we get:

$$P \left( \frac{1}{T^2} \right)^{\gamma} = \text{constant}$$.

Simplifying, we have:

$$P^{1-\gamma} T^{2\gamma} = \text{constant}$$.

Since P is proportional to $$T^3$$, we can write:

$$(T^3)^{1-\gamma} T^{2\gamma} = \text{constant}$$.

This simplifies to:

$$T^{3-3\gamma + 2\gamma} = \text{constant}$$.

For this equation to hold, the exponent of T must be zero. Therefore:

$$3 - 3\gamma + 2\gamma = 0$$.

Solving for $$\gamma$$, we get:

$$\gamma = \frac{C_p}{C_v} = \boxed{\frac{3}{2}}$$.

Therefore, the correct answer is Option B.