The answer is B. For 60 residents, $\operatorname{since} \sin \frac{\pi}{60}>\frac{1}{10} \sin \frac{\pi}{6}=\frac{1}{20}$, the side length of a regular 60 -gon inscribed in $C$ is not less than 1 meter. Therefore, $P_{1}, P_{2}, \ldots, P_{60}$ may be all vertices of this polygon. For different $i, j$, the distance from $P_{i}$ to $K P_{j}$ is not less than $10 \sin \frac{\pi}{30}$ meter. Since $\sin \frac{\pi}{30}>\frac{1}{5} \sin \frac{\pi}{6}=\frac{1}{10}$, all residents could see the entire show. On the other hand, if there are 800 residents, we draw rays $\overrightarrow{K P_{1}}, \overrightarrow{K P_{2}}, \ldots, \overrightarrow{K P_{800}}$. If two of them (called $\overrightarrow{K P}_{i}, \overrightarrow{K P}_{j}$ ) coincide, and $K P_{j}>K P_{i}$, then $A_{j}$ 's sight line is blocked by $A_{i}$. Suppose no two rays coincide, we first prove that if an angle $\angle P_{i} K P_{j}$ satisfies $\angle P_{i} K P_{j} \leq$ $\frac{1}{12}\left(\frac{1}{K P_{i}}+\frac{1}{K P_{j}}\right)$ (the unit of angle is rad, and the unit of length is meter), then the sight line of one of $A_{i}, A_{j}$ is blocked by the other. Without loss of generality we suppose $K P_{i} \leq K P_{j}$. Since $K P_{i}, K P_{j} \geq 10$, we get that $\angle P_{i} K P_{j} \leq \frac{1}{60}$, so it is acute. Therefore, the foot point from $P_{i}$ to $K P_{j}$ is inside segment $K P_{j}$, and the distance from $P_{i}$ to $K P_{j}$ is $$K P_{i} \sin \angle P_{i} K P_{j}<K P_{i} \cdot \angle P_{i} K P_{j} \leq \frac{K P_{i}}{12}\left(\frac{1}{K P_{i}}+\frac{1}{K P_{j}}\right) \leq \frac{1}{6}$$ So, $A_{j}$ 's sight line is blocked by $A_{i}$. Note that $\overrightarrow{K P_{1}}, \overrightarrow{K P_{2}}, \ldots, \overrightarrow{K P_{800}}$ cut the perigon with vertex $K$ to 800 angles, and their sum is $2 \pi$, but the sum of all $\frac{1}{12}\left(\frac{1}{K P_{i}}+\frac{1}{K P_{j}}\right)$ is $$\begin{aligned} \sum \frac{1}{12}\left(\frac{1}{K P_{i}}+\frac{1}{K P_{j}}\right) & =\frac{1}{6} \sum_{i=1}^{800} \frac{1}{K P_{i}} \\ & \geq \frac{1}{6} \sum_{i=1}^{800} \frac{1}{\sqrt{i+99}} \text { (from the conclusion of (1)) } \\ & =\frac{1}{6} \sum_{m=100}^{899} \frac{1}{\sqrt{m}} \\ & \geq \frac{1}{6} \sum_{m=100}^{899}\left(\int_{m}^{m+1} \frac{1}{\sqrt{x}} d x\right) \quad \text { (since } \frac{1}{\sqrt{x}} \text { is a decreasing function) } \\ & =\frac{1}{6} \int_{100}^{900} \frac{1}{\sqrt{x}} d x=\frac{\sqrt{900}-\sqrt{100}}{3}=\frac{20}{3}>2 \pi \end{aligned}$$ Therefore, there exist an angle $\angle P_{i} K P_{j}$ satisfies $\angle P_{i} K P_{j} \leq \frac{1}{12}\left(\frac{1}{K P_{i}}+\frac{1}{K P_{j}}\right)$, and some of the residents could not see the entire show.