First, the intensity $$I$$ at a distance $$r$$ from a point light source spreading its energy isotropically is given by

$$I = \frac{P}{4\pi r^2},$$

where

$$P = 450 \, \text{W}$$

$$r = 2 \, \text{m}.$$

So,

$$I = \frac{450}{4\pi (2)^2} = \frac{450}{16\pi} \approx \frac{450}{50.27} \approx 8.96 \, \text{W/m}^2.$$

For a perfectly reflecting surface, the radiation pressure $$p$$ is given by

$$p = \frac{2I}{c},$$

where $$c \approx 3 \times 10^8 \, \text{m/s}$$ is the speed of light.

Substituting the value of $$I$$:

$$p = \frac{2 \times 8.96}{3 \times 10^8} \approx \frac{17.92}{3 \times 10^8} \, \text{Pa}.$$

Calculating the final value:

$$p \approx 5.97 \times 10^{-8} \, \text{Pa},$$

which is approximately $$6 \times 10^{-8} \, \text{Pa}.$$

Thus, the correct answer is:

Option D: $$6 \times 10^{-8}$$ Pascals.