\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2018}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\left(a+b+c=2018\right)\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right]\left(a+b\right)=0\)
\(\Leftrightarrow\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}\times\left(a+b\right)=0\)
\(\Leftrightarrow\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\b=-c\\a=-b\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}b=2018\\a=2018\\c=2018\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{1}{2018^{2017}}\)
