1. Distinct complex numbers $a, b, c$ satisfy $abc=1$, $ab+a+1 \neq 0$, and let
$$
\begin{array}{l}
M=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1} \\
N=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}, \\
P=\frac{ab}{ab+a+1}+\frac{bc}{bc+b+1}+\frac{ca}{ca+c+1} .
\end{array}
$$

Then the relationship between $M, N, P$ is ( ).
(A) $M=N \neq P$
(B) $M=P \neq N$
(C) $N=P \neq M$
(D) $M=N=P$
- $1 .(\mathrm{D})$.
$$
\begin{array}{l}
\text { From } \frac{1}{a b+a+1}=\frac{b c}{b c(a b+a+1)}=\frac{b c}{b+1+b c}, \\
\frac{1}{a b+a+1}=\frac{c}{c(a b+a+1)}=\frac{c}{1+a c+c},
\end{array}
$$

we get $\frac{1}{a b+a+1}=\frac{b c}{b c+b+1}=\frac{c}{a c+c+1}$.
Similarly, $\frac{a}{a b+a+1}=\frac{1}{b c+b+1}=\frac{a c}{a c+c+1}$,
$$
\frac{a b}{a b+a+1}=\frac{b}{b c+b+1}=\frac{1}{a c+c+1} \text {. }
$$

Adding them up, we get $M=N=P$.
Another proof hint: Calculate the values of $M, N, P$ separately or calculate
$$
M=N, M=P \text {. }
$$