To begin with, we're told that during an adiabatic process, the pressure $P$ of a gas is directly proportional to the cube of its absolute temperature $T$, that is, $P \propto T^3$. From this, we can express the relation as $P = kT^3$, where $k$ is a constant.

In an adiabatic process, $PV^\gamma = \text{constant}$, where $\gamma = \frac{C_P}{C_V}$, $P$ is the pressure, $V$ is the volume, $C_P$ is the heat capacity at constant pressure, and $C_V$ is the heat capacity at constant volume.

Also, the ideal gas equation is $PV = nRT$, where $n$ is the amount of substance, $R$ is the ideal gas constant, and $T$ is the absolute temperature. This can be rearranged to express $P$ in terms of $T$ and $V$, resulting in $P = \frac{nRT}{V}$.

Given that $P = kT^3$, we can equate this to the expression we got from the ideal gas law: $\frac{nRT}{V} = kT^3$. Simplifying, we get $V = \frac{nR}{kT^2}$, which shows that $V$ is inversely proportional to $T^2$.

Now, let's recall the adiabatic condition $PV^\gamma = \text{constant}$. Substituting the proportionalities $P \propto T^3$ and $V \propto T^{-2}$ into this equation, it becomes $T^3(T^{-2})^\gamma = \text{constant}$, simplifying to $T^{3-2\gamma} = \text{constant}$.

For the expression $T^{3-2\gamma} = \text{constant}$ to hold true in an adiabatic process, where the only variable is temperature $T$, the exponent must equal to zero (as the quantity of $T$ to some power equals to a constant suggests that the change in $T$ does not alter the value of the expression). This implies $3 - 2\gamma = 0$, solving for $\gamma$:

$3 - 2\gamma = 0 \implies 2\gamma = 3 \implies \gamma = \frac{3}{2}$

Therefore, the ratio of $\frac{C_P}{C_V}$ for the gas is $\frac{3}{2}$, which matches with Option B $\frac{3}{2}$.