Write the given condition in the form
$$x + y = 3(\sqrt{x+1} + \sqrt{y+2}).$$
Denote by $G$ the set of values of $P$.
It is easily seen that:
$a \in G \Leftrightarrow a$ is a real number so that the following system of equations (with unknowns $x, y$) has solutions
$$\begin{cases} 3(\sqrt{x+1} + \sqrt{y+2}) = a \\ x+y = a. \end{cases} \qquad (I)$$
By putting $u = \sqrt{x+1}$ and $v = \sqrt{y+2}$, from the system (I), we get the following system of equations (with unknowns $u, v$)
$$\begin{cases} 3(u+v) = a \\ u^2 + v^2 = a+3. \end{cases} \qquad (II)$$
which is equivalent to
$$\begin{cases} u+v = \frac{a}{3} \\ uv = \frac{1}{2} \left( \frac{a^2}{9} - a - 3 \right). \end{cases}$$
Therefore,
The system (I) has solutions $\Leftrightarrow$ the system (II) has the solutions $(u, v)$ with $u, v$ non negative $\Leftrightarrow$ the equation (with unknown $t$) $18t^2 - 6at + a^2 - 9a - 27 = 0$ has two non negative roots $\Leftrightarrow$
$$\begin{cases} -a^2 + 18a + 54 \ge 0 \\ a \ge 0 \\ a^2 - 9a - 27 \ge 0 \end{cases} \Leftrightarrow \frac{9 + 3\sqrt{21}}{2} \le a \le 9 + 3\sqrt{15}.$$
Hence $G = \left[ \frac{9 + 3\sqrt{21}}{2} ; 9 + 3\sqrt{15} \right]$ and so $\min P = \frac{9 + 3\sqrt{21}}{2}$, $\max P = 9 + 3\sqrt{15}$.