1. Let $p>3, p$ and $p+4$ are both prime numbers, denote $S_{1}=p(p+$ 4), for $i=1,2, \cdots, k-1, S_{i+1}$ represents the sum of the digits of $S_{i}$. If $S_{\star}$ is a single digit, then $S_{k}=(\quad)$.
(A) 3
(B) 5
(C) 7
(D) 9
-1. (13)
If $p \equiv (\bmod 3)$, then $p+4 \equiv 0(\bmod 3)$, which contradicts that $p+4$ is a prime number. Therefore, $p \equiv 1(\bmod 3)$, so $3 \mid (p+2)$.
$$
\text { Also, } \begin{aligned}
S_{1} & =p(p+4)=p^{2}+4 p+4-4 \\
& =(p+2)^{2}-9+5,
\end{aligned}
$$

then $p(p+4) \equiv 5(\bmod 9)$. Thus, $S_{k}=5$.