The reaction is:

$$\mathrm{CO}_2(g) + \mathrm{C}(s) \rightleftharpoons 2\mathrm{CO}(g)$$

Initially, the pressure of $\mathrm{CO}_2$ is 0.5 atm, and the pressure of $\mathrm{CO}$ is 0 atm. Let 'x' be the change in pressure of $\mathrm{CO}_2$ at equilibrium. Since the stoichiometric coefficient of CO is twice that of $\mathrm{CO}_2$, the pressure of CO formed at equilibrium is 2x.

At equilibrium:

Pressure of $\mathrm{CO}_2 = 0.5 - x$

Pressure of $\mathrm{CO} = 2x$

The total pressure at equilibrium is given as 0.8 atm. Therefore:

$$(0.5 - x) + 2x = 0.8$$

$$0.5 + x = 0.8$$

$$x = 0.8 - 0.5 = 0.3$$

So, at equilibrium:

Pressure of $\mathrm{CO}_2 = 0.5 - 0.3 = 0.2 \text{ atm}$

Pressure of $\mathrm{CO} = 2(0.3) = 0.6 \text{ atm}$

The equilibrium constant $K_p$ is given by:

$$K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}$$

$$K_p = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \text{ atm}$$

Therefore, the value of $K_p$ is 1.8 atm.