Infinite!

$\begin{aligned} P & = \prod_{k=2}^\infty \frac{k^3-1}{k^3+1} \\ & = \prod_{k=2}^\infty \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} \\ & = \frac{\displaystyle \prod_{k=2}^\infty (k-1) \prod_{k=2}^\infty (k^2+k+1)}{\displaystyle \prod_{k=2}^\infty (k+1) \prod_{k=2}^\infty (k^2-k+1)} \\ & = \frac{\displaystyle \prod_{k=1}^\infty k \times \prod_{k=2}^\infty (k^2+k+1)}{\displaystyle \prod_{k=3}^\infty k \times \prod_{k=\color{#3D99F6}{1}}^\infty (\color{#3D99F6}{k^2+k+1})} \quad \quad \small \color{#3D99F6}{\text{We note that }(k+1)^2-(k+1)+1 = k^2 + k +1} \\ & = \frac{1\times 2}{1^2+1+1} = \frac{2}{3} \approx \boxed{0.67} \end{aligned}$