The radiation pressure exerted on the metal plate can be calculated using the formula:

$$P = \frac{2I}{c}$$

where $$I$$ is the intensity of the incident radiation and $$c$$ is the speed of light.

Given that the metal plate absorbs 80% of the incident radiation with a power of 500 W, the intensity of the incident radiation can be calculated as:

$$I = \frac{0.8 \cdot 500}{0.5} = 800 \text{ W/m}^2$$

Substituting this value into the formula for radiation pressure, we get:

$$P = \frac{2 \cdot 800}{3 \cdot 10^8} = 5.33 \cdot 10^{-6} \text{ N/m}^2$$

Therefore, the radiation pressure exerted on the metal plate is approximately [1] $$5.33 \cdot 10^{-6} \text{ N/m}^2$$