To determine the final pressure of the gas after undergoing an isothermal expansion followed by an adiabatic compression, let's break down the problem step-by-step.

Step 1: Isothermal Expansion

During an isothermal process, the temperature remains constant. For an ideal gas, the relationship between pressure and volume is given by:

$$P_1 V_1 = P_2 V_2$$

Here, the initial pressure is $$\mathrm{P}$$ and the initial volume is $$\mathrm{V}$$. After the isothermal expansion, the volume becomes $$9 \mathrm{~V}$$ and the pressure becomes $$\mathrm{P_2}$$.

So we have:

$$\mathrm{P} \cdot \mathrm{V} = \mathrm{P_2} \cdot 9 \mathrm{V}$$

Solving for $$\mathrm{P_2}$$:

$$\mathrm{P_2} = \frac{\mathrm{P}}{9}$$

Step 2: Adiabatic Compression

During an adiabatic process, no heat is exchanged with the surroundings. The relationship between pressure and volume for an adiabatic process is given by:

$$P_2 V_2^\gamma = P_3 V_3^\gamma$$

Here, $$\gamma$$ is the adiabatic index, which is the ratio of specific heat at constant pressure to constant volume ($$=\frac{3}{2} = 1.5$$). The final volume $$V_3$$ is equal to the initial volume $$\mathrm{V}$$.

Given:

$$\mathrm{P_2} = \frac{\mathrm{P}}{9}, \quad \mathrm{V_2} = 9 \mathrm{V}, \quad \mathrm{V_3} = \mathrm{V}$$

Thus, we have:

$$\left(\frac{\mathrm{P}}{9}\right) \cdot (9 \mathrm{V})^\gamma = \mathrm{P_3} \cdot \mathrm{V}^\gamma$$

Substituting $$\gamma = 1.5$$:

$$\left(\frac{\mathrm{P}}{9}\right) \cdot (9 \mathrm{V})^{1.5} = \mathrm{P_3} \cdot \mathrm{V}^{1.5}$$

Since $$\gamma = 1.5$$, we have:

$$ (9 \mathrm{V})^{1.5} = 9^{1.5} \cdot \mathrm{V}^{1.5} = 27 \cdot \mathrm{V}^{1.5}$$

Thus:

$$ \left(\frac{\mathrm{P}}{9}\right) \cdot 27 \cdot \mathrm{V}^{1.5} = \mathrm{P_3} \cdot \mathrm{V}^{1.5} $$

Dividing both sides by $$\mathrm{V}^{1.5}$$:

$$ \left(\frac{\mathrm{P}}{9}\right) \cdot 27 = \mathrm{P_3} $$

Simplifying, we get:

$$\mathrm{P_3} = 3 \mathrm{P}$$

So, the final pressure of the gas after the adiabatic compression is 3P.

Therefore, the correct answer is:

Option C: 3 P

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