Here, $\begin{align}P =\lim_{n\to \infty}\prod_{r=2}^{n} \frac{r^3-1}{r^3+1}&=\lim_{n\to \infty}\prod_{r=2}^{n} \left(\frac{r-1}{r+1}\right)\left(\frac{r^2+r+1}{r^2-r+1}\right)\\&=\lim_{n\to \infty}\left(\prod_{r=2}^{n} \frac{r-1}{r+1}\right) \left( \prod_{r=2}^{n} \frac{r^2+r+1}{r^2-r+1}\right)......\tag{i}\end{align}$
Now
$\begin{align} \prod_{r=2}^{n} \frac{r-1}{r+1}&=\frac{1}{3}\times\frac{2}{4}\times\frac{3}{5}\times\frac{4}{6}\times\frac{5}{7}\times\frac{6}{8}\times\cdots\times\frac{n-2}{n}\times\frac{n-1}{n+1}\\&=\frac{1\times2}{n(n+1)}; [\scriptsize \text{All numbers, except the first two numerators and the last two denominators,}\\&~~~~~~~~~~~~~~~~~~~~~~\scriptsize\text{ are crossed out by each other.}]\\&=\frac{2}{n(n+1)} \end{align}$
Then
$\begin{align} &~~~~\prod_{r=2}^{n} \frac{r^2+r+1}{r^2-r+1}\\&=\prod_{r=2}^{n} \frac{r(r+1)+1}{r(r-1)+1}\\&=\scriptsize\frac{(2\times3+1)}{(2\times1+1)}\times\frac{(3\times4+1)}{(3\times2+1)}\times\frac{(4\times5+1)}{(4\times3+1)}\times\cdots\times\frac{(n-1)n+1}{(n-1)(n-2)+1}\times\frac{n(n+1)+1}{n(n-1)+1}\\&=\frac{n(n+1)+1}{2\times1+1}; [\scriptsize \text{All numbers, except the first denominator and the last numerator,}\\&~~~~~~~~~~~~~~~~~~~~~~~~~~~~\scriptsize\text{ are crossed out by each other.}]\\&=\frac{n(n+1)+1}{3} \end{align}$
Now putting these values to no$\mathrm{(i)}$, we get
$$\begin{align}P &=\lim_{n\to \infty}\left(\frac{2}{n(n+1)}\right) \left( \frac{n(n+1)+1}{3}\right)\\&=\frac{2}{3}\lim_{n\to \infty} \left( 1+\frac{1}{n(n+1)} \right)\\&=\frac{2}{3}; ~~[\because n\to \infty \Rightarrow \frac{1}{n}\to 0 \Rightarrow \frac{1}{n(n+1)}\to 0] \end{align}$$
So the correct answer is B.