Ta có:

\[\begin{array}{l}
{P^2} = \frac{{{x^2}{y^2}}}{{{z^2}}} + \frac{{{y^2}{z^2}}}{{{x^2}}} + \frac{{{z^2}{x^2}}}{{{y^2}}} + 2{x^2} + 2{y^2} + 2{z^2}\\
2{P^2} = 2\frac{{{x^2}{y^2}}}{{{z^2}}} + 2\frac{{{y^2}{z^2}}}{{{x^2}}} + 2\frac{{{z^2}{x^2}}}{{{y^2}}} + 4{x^2} + 4{y^2} + 4{z^2}
\end{array}\]

Theo bất đẳng thức Cosi ta có:

\[\begin{array}{l}
\frac{{{x^2}{y^2}}}{{{z^2}}} + \frac{{{y^2}{z^2}}}{{{x^2}}} \ge 2\sqrt {\frac{{{x^2}{y^2}}}{{{z^2}}}.\frac{{{y^2}{z^2}}}{{{x^2}}}} = 2{y^2}\\
\frac{{{y^2}{z^2}}}{{{x^2}}} + \frac{{{z^2}{x^2}}}{{{y^2}}} \ge 2\sqrt {\frac{{{y^2}{z^2}}}{{{x^2}}}.\frac{{{z^2}{x^2}}}{{{y^2}}}} = 2{z^2}\\
\frac{{{x^2}{y^2}}}{{{z^2}}} + \frac{{{z^2}{x^2}}}{{{y^2}}} \ge 2\sqrt {\frac{{{x^2}{y^2}}}{{{z^2}}}.\frac{{{z^2}{x^2}}}{{{y^2}}}} = 2{x^2}
\end{array}\]

Do đó:

\[\begin{array}{l}
\Rightarrow 2{P^2} = 2\frac{{{x^2}{y^2}}}{{{z^2}}} + 2\frac{{{y^2}{z^2}}}{{{x^2}}} + 2\frac{{{z^2}{x^2}}}{{{y^2}}} + 4{x^2} + 4{y^2} + 4{z^2} \ge 2{y^2} + 2{z^2} + 2{x^2} + 4{x^2} + 4{y^2} + 4{z^2}\\
\Leftrightarrow {P^2} \ge 3\left( {{x^2} + {y^2} + {z^2}} \right) = 3.2019 = 6057\\
\Leftrightarrow P \ge \sqrt {6057} \left( {P > 0} \right) \Leftrightarrow {P_{\min }} = \sqrt {6057} \Leftrightarrow x = y = z = \sqrt {\frac{{2019}}{3}}
\end{array}\]

Vậy:

\[{P_{\min }} = \sqrt {6057} \]