Answer
$$K_p = \frac{P_{COCl_2}^2}{P_{CO_2}P_{Cl_2}^2}$$
$K_p = 4.7 \times 10^9$
Work Step by Step
1. Multiply the second reaction by 2; to do so, raise $K_p^{''}$ to the power of 2.
$$2 CO(g) + 2Cl_2(g) \leftrightharpoons 2COCl_2(g)$$
$K_p^{''} = (6.0 \times 10^{-3})^2 = 3.6 \times 10^{-5}$
2. Add both reactions; removing the repeated components, the Kp of the resulting reaction is the multiplication of $K_p^{'}$ and the new $K_p^{''}$
$$C(s) + CO_2(g) + 2Cl_2(g) \leftrightharpoons 2COCl_2(g)$$
$K_p = 1.3 \times 10^{14} \times 3.6 \times 10^{-5} = 4.7 \times 10^9$
3. Write the equilibrium constant expression:
$$K_p = \frac{[Products]}{[Reactants]} = \frac{P_{COCl_2}^2}{P_{CO_2}P_{Cl_2}^2}$$
