$P+\left ( \frac{1}{a}+\frac{1}{b} +\frac{1}{c} \right )=\dfrac{a-1+b-1}{b^2}+\dfrac{b-1+c-1}{c^2}+\dfrac{c-1+a-1}{a^2}$
$=(a-1)\left ( \frac{1}{a^2}+\frac{1}{b^2} \right )+(b-1)\left ( \frac{1}{b^2}+\frac{1}{c^2} \right )+(c-1)\left ( \frac{1}{c^2}+\frac{1}{a^2} \right )$
$\ge (a-1) \frac{2}{ab}+(b-1)\frac{2}{bc}+(c-1)\frac{2}{ac}$
$=\frac{2}{a}+\frac{2}{b} +\frac{2}{c} -\left ( \frac{1}{ab}+\frac{1}{bc} +\frac{1}{ca} \right )$
$\implies P \ge \frac{1}{a}+\frac{1}{b} +\frac{1}{c}-\frac{a+b+c}{abc}$
$\ge \sqrt{3\left ( \frac{1}{ab}+\frac{1}{bc} +\frac{1}{ca} \right )}-1=\sqrt 3-1$.
Vậy $\min P=\sqrt 3-1\Leftrightarrow a=b=c=\sqrt 3.$