**Problem Analysis:**
We are given that the initial volume of the neon gas sample is decreased by 15% by cooling the gas by 319 degrees Celsius. We need to find the temperature to which the gas must be heated to double its pressure. To solve this problem, we can use the ideal gas law equation, which relates the pressure, volume, and temperature of a gas.

**Solution:**

1. Let's assume the initial volume of the gas sample is V1 and the initial temperature is T1. The initial pressure can be calculated using the ideal gas law equation:
P1 = (nRT1)/V1 (where n is the number of moles of the gas and R is the ideal gas constant)

2. After cooling the gas, the volume decreases by 15%, so the new volume becomes V2 = 0.85V1.

3. The temperature also changes due to cooling, and the new temperature becomes T2 = T1 - 319 degrees Celsius.

4. We are given that we need to find the temperature at which the pressure doubles. Let's assume this temperature is T3.

5. Using the ideal gas law equation again, we can calculate the new pressure at temperature T2:
P2 = (nRT2)/V2

6. We can also calculate the pressure at temperature T3, which is double the initial pressure:
P3 = 2P1 = 2(nRT1)/V1

7. We can set P2 equal to P3 and solve for T3. Rearranging the equation, we get:
(nRT2)/V2 = 2(nRT1)/V1

8. Since the number of moles (n) and the ideal gas constant (R) are the same on both sides of the equation, we can cancel them out:
(T2/V2) = 2(T1/V1)

9. Substituting the values of T2, V2, T1, and V1, we get:
(T1 - 319)/0.85V1 = 2(T1/V1)

10. Solving for T1, we get:
T1 - 319 = 1.7T1
0.7T1 = 319
T1 = 319/0.7
T1 ≈ 455.714 degrees Celsius

11. Therefore, the temperature to which the neon gas sample must be heated to double its pressure is approximately 455.714 degrees Celsius.

**Answer:**
The correct option is **b. 592 degrees Celsius**.