[Solution] Obviously, within a certain specific range, the terms in $P$ containing $x$ will be eliminated, $\because 2+3+\cdots+10=54$, it will eventually form: $27 x-27 x=0$, since $8 x+9 x+10 x=27 x$, it is easy to see:
When $\frac{1}{8}<x<\frac{1}{7}$,
$$
\begin{aligned}
P= & 1-2 x+1-3 x+1-4 x+1-5 x+1-6 x+1-7 x \\
& +8 x-1+9 x-1+10 x-1 \\
= & 3+(8 x+9 x+10 x-2 x-3 x-4 x-5 x-6 x-7 x) \\
= & 3 .
\end{aligned}
$$

Therefore, the answer is $(B)$.
1.57 If for any allowed value of $x$ within a certain range, $P=|1-2 x|+$ $|1-3 x|+|1-4 x|+\cdots+|1-9 x|+|1-10 x|$ is a constant, then the constant is
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(Hefei, Anhui Province, China Junior High School Mathematics Competition, 1990)