II. (50 points)
Let $a, b, c \in \mathbf{R}^{+}$, and $a b c + a + c = b$. Determine the maximum value of $P = \frac{2}{a^{2}+1} - \frac{2}{b^{2}+1} + \frac{3}{c^{2}+1}$.
Given the conditions, we have \(a + c = (1 - ac)b\) and \(1 - ac \neq 0\) (otherwise, \(a + c = 0\), which is impossible for \(a, c \in \mathbf{R}'\)).

Thus, \(b = \frac{a + c}{1 - ac}\). Let \(\alpha = \arctan a\), \(\beta = \arctan b\), \(\gamma = \arctan c\), where \(\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)\). Then, \(\tan \beta = \frac{\tan \alpha + \tan \gamma}{1 - \tan \alpha \tan \gamma} = \tan (\alpha + \gamma)\). Since \(\beta, \alpha + \gamma \in (0, \pi)\), we have \(\beta = \alpha + \gamma\). Therefore,
\[
\begin{aligned}
P & = \frac{2}{1 + \tan^2 \alpha} - \frac{2}{1 + \tan^2 \beta} + \frac{3}{1 + \tan^2 \gamma} \\
& = 2 \cos^2 \alpha - 2 \cos^2 (\alpha + \gamma) + 3 \cos^2 \gamma \\
& = (\cos 2\alpha + 1) - [\cos (2\alpha + 2\gamma) + 1] + 3 \cos^2 \gamma \\
& = \cos 2\alpha - \cos (2\alpha + 2\gamma) + 3 \cos^2 \gamma = 2 \sin \gamma \sin (2\alpha + \gamma) + 3 \cos^2 \gamma \\
& \leq 2 \sin \gamma + 3 \cos^2 \gamma = \frac{10}{3} - 3 \left(\sin \gamma - \frac{1}{3}\right)^2 \leq \frac{10}{3}
\end{aligned}
\]

The equality holds when \(2\alpha + \gamma = \frac{\pi}{2}\) and \(\sin \gamma = \frac{1}{3}\), i.e., \(a = \frac{\sqrt{2}}{2}\), \(b = \sqrt{2}\), \(c = \frac{\sqrt{2}}{4}\). Therefore, \(P_{\max} = \frac{10}{3}\).