4. For a finite sequence $P=\left(p_{1}, p_{2}, \cdots, p_{n}\right)$, the Cesàro sum of $P$ (Cesàro being a mathematician's name) is defined as $\frac{1}{n}\left(S_{1}+S_{2}+\cdots+S_{n}\right)$, where $S_{k}=p_{1}+p_{2}+\cdots+p_{k}(1 \leqslant k \leqslant n)$. If a 99-term sequence $\left(p_{1}, p_{2}, \cdots, p_{99}\right)$ has a Cesàro sum of 1000, then the Cesàro sum of the 100-term sequence $\left(1, p_{1}, p_{2}, \cdots, p_{99}\right)$ is ( ).
A. 990
B. 991
C. 992
D. 993
4. B,
$$
\begin{aligned}
\frac{1}{99}\left(S_{1}+S_{2}+\cdots+S_{99}\right) & =1000, S_{1}+S_{2}+\cdots+S_{99}=99000 \text {, so } \frac{1}{100}\left[1+\left(1+S_{1}\right)+\right. \\
\left.\left(1+S_{2}\right)+\cdots+\left(1+S_{99}\right)\right] & =\frac{1}{100}\left[100+\left(S_{1}+S_{2}+\cdots+S_{99}\right)\right]=\frac{1}{100}[100+99000]=991 .
\end{aligned}
$$