l energy, $$\Delta U$$, for an ideal gas can be calculated using the specific heats and the first law of thermodynamics. It's given by the equation:

$$\Delta U = n C_v \Delta T$$

where $$\Delta U$$ is the change in internal energy, $$n$$ is the amount of substance, $$C_v$$ is the molar specific heat capacity at constant volume, and $$\Delta T$$ is the change in temperature.

Since we're dealing with a process at constant pressure, we can also relate the change in temperature to the change in volume using the ideal gas law. Under constant pressure, the change in volume $$\Delta V$$ and the change in temperature $$\Delta T$$ are related by the equation:

$$P \Delta V = n R \Delta T$$

Given that the volume changes from $$V$$ to $$3V$$, we have:

$$\Delta V = 3V - V = 2V$$

Substituting this into the equation gives us:

$$P \cdot 2V = n R \Delta T$$

Now, relating the specific heats, we know that:

$$\gamma = \frac{C_p}{C_v}$$

where $$\gamma$$ is the given ratio of specific heats, $$C_p$$ is the molar specific heat at constant pressure, and $$C_v$$ is the molar specific heat at constant Volume. The molar specific heat capacities are related to the gas constant $$R$$ by:

$$C_p - C_v = R$$

Substituting $$\gamma = \frac{C_p}{C_v}$$ into the expression and solving for $$C_v$$ gives:

$$C_v = \frac{R}{\gamma - 1}$$

Therefore, the change in internal energy can be also expressed in terms of $$\gamma$$, $$R$$, and $$\Delta T$$:

$$\Delta U = n \left(\frac{R}{\gamma - 1}\right) \Delta T = \frac{P \cdot 2V}{\gamma - 1}$$

So, the change in internal energy when the volume changes from $$V$$ to $$3V$$ at constant pressure is $$\frac{2 P V}{(\gamma-1)}$$, which is Option D.

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