\(a+b+c=1\Rightarrow\hept{\begin{cases}ab+c=ab+c\left(a+b+c\right)\\bc+a=bc+a\left(a+b+c\right)\\ca+b=ca+b\left(a+b+c\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=ab+ca+bc+c^2\\bc+a=bc+a^2+ab+ac\\ca+b=ca+ab+b^2+bc\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}ab+c=\left(b+c\right)\left(a+c\right)\\bc+a=\left(a+c\right)\left(a+b\right)\\ca+b=\left(b+c\right)\left(a+b\right)\end{cases}}\)
\(\Rightarrow P=\frac{\left(b+c\right)\left(a+c\right)}{\left(a+b\right)^2}.\frac{\left(a+c\right)\left(a+b\right)}{\left(b+c\right)^2}.\frac{\left(b+c\right)\left(a+b\right)}{\left(c+a\right)^2}=1\)