No decrease of generality, assume$x\leq y\leq z$
Set $t=x+y+z$
From the assumptions, applied inequality Cauchy, we have:
$7=\frac{x+y}{z}+z(\frac{1}{x}+\frac{1}{y})+\frac{x}{y}+\frac{y}{x}\geq \frac{2\sqrt{xy}}{z}+\frac{2z}{\sqrt{xy}}+2$
So that:
$2\sqrt{xy}\geq z\rightarrow x+y\geq z$
$\Rightarrow (x+y-z)(y+z-x)(z+x-y)\geq 0$
$\Leftrightarrow x^3+y^2+z^3\leq \sum xy(x+y)-2xyz$
Assumptions rewritten as:
$(x+y+z)(xy+yz+zx)=10xyz$
$\rightarrow $ We have:
$\sum xy(x+y)=7xyz$
and:
$\prod (x+y)=9xyz$
So that:
$x^3+y^3+z^3\leq \sum xy(x+y)-2xyz=5xyz$
We always have:
$x^3+y^3+z^3+3\prod (x+y)=(x+y+z)^3$
$\rightarrow x^3+y^3+z^3\leq \frac{5}{9}\prod (x+y)\leq \frac{5}{32}t^3$
and:
$xy+yz+zx=\frac{10xyz}{x+y+z}\geq \frac{5}{16}.\frac{x^3+y^3+z^3+3\prod (x+y)}{t}=\frac{5}{16}t^2$
So:
$P\leq \frac{48}{5t^2}-\frac{128}{5t^3}=f(t)$
We have:
$f'(t)=\frac{96(4-t)}{5t^4}$
Tabulation of change of $f (t)$ we get $f(t)\leq f(4)=\frac{1}{5}$
$\rightarrow P\leq \frac{1}{5}$ at $x=y=1;z=2./$