$$2$$ SO$$_2$$(g) + O$$_2$$(g) $$\rightleftharpoons$$ 2 SO$$_3$$(g)
In an equilibrium mixture, the partial pressures are
P$$_{SO_3}$$ = 43 kPa; P$$_{O_2}$$ = 530 Pa and
P$$_{SO_2}$$ = 45 kPa. The equilibrium constant K$$_P$$ = ___ $$\times 10^{-2}$$. (Nearest integer)

For the equilibrium $$2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})$$, the equilibrium constant $$K_P$$ is defined as:

$$K_P = \frac{(P_{\text{SO}_3})^2}{(P_{\text{SO}_2})^2 \cdot P_{\text{O}_2}}$$

The given partial pressures are: $$P_{\text{SO}_3} = 43 \text{ kPa}$$, $$P_{\text{O}_2} = 530 \text{ Pa} = 0.530 \text{ kPa}$$, and $$P_{\text{SO}_2} = 45 \text{ kPa}$$. All pressures must be in consistent units (kPa).

Substituting:

$$K_P = \frac{(43)^2}{(45)^2 \times 0.530} = \frac{1849}{2025 \times 0.530} = \frac{1849}{1073.25}$$

$$K_P = \frac{1849}{1073.25} \approx 1.7228 \text{ kPa}^{-1}$$

Expressing as $$K_P = p \times 10^{-2}$$: $$p = 1.7228 \times 10^{2} \times 10^{-2} = 172.28$$

Rounding to the nearest integer, $$K_P \approx 172 \times 10^{-2} \text{ kPa}^{-1}$$, so the answer is $$\boxed{172}$$.