Ta có : $\dfrac{2}{xy} - 9 = \dfrac{1}{z^2}$
$\to 9 = \dfrac{2}{xy} - \dfrac{1}{z^2}$
Lại có : $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} = 3$
$\to \bigg(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\bigg)^2 = 9$
$\to \dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2.\bigg(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\bigg) = \dfrac{2}{xy} - \dfrac{1}{z^2}$
$\to \dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{z^2}+\dfrac{2}{yz}+\dfrac{2}{zx} = 0 $
$\to \bigg(\dfrac{1}{x}+\dfrac{1}{z}\bigg)^2 + \bigg(\dfrac{1}{y}+\dfrac{1}{z}\bigg)^2 = 0 $
Dấu "=" xảy ra $⇔x=y=-z$
Kết hợp với giả thiết thì $x=y=\dfrac{1}{3}, z= -\dfrac{1}{3}$
Khi đó $P = \bigg(\dfrac{1}{3}+3.\dfrac{1}{3}-\dfrac{1}{3}\bigg)^{2019} = 1$
Vậy $P=1$
