1. Let $a, b, c, p$ be real numbers, not all equal, and
$$
a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}=p \text {. }
$$

Then $a^{2} b^{2} c^{2}=(\quad)$.
(A) $-p$
(B) $p$
(C) $p^{2}$
(D) $\sqrt{p}$
- 1. C.

Given $p=c+\frac{p-\frac{1}{c}}{\left(p-\frac{1}{c}\right) p-1}$. Simplifying, we get $\left(p^{2}-1\right)\left(p-c-\frac{1}{c}\right)=0$. When $p=c+\frac{1}{c}$, we have $a=b=c$, which is a contradiction.
Therefore, $p^{2}=1$.
From the given information,
$$
\begin{array}{l}
a+\frac{1}{b}=b+\frac{1}{c} \Rightarrow a-b=\frac{b-c}{b c} \\
\Rightarrow b c=\frac{b-c}{a-b} .
\end{array}
$$

Similarly, $b+\frac{1}{c}=c+\frac{1}{a} \Rightarrow c a=\frac{c-a}{b-c}$, $c+\frac{1}{a}=a+\frac{1}{b} \Rightarrow a b=\frac{a-b}{c-a}$.
Thus, $a^{2} b^{2} c^{2}=\frac{b-c}{a-b} \cdot \frac{c-a}{b-c} \cdot \frac{a-b}{c-a}=1=p^{2}$.