The correct answer is $\boxed{\text{B. N – 1}}$.
The rank of a matrix is the number of linearly independent rows or columns in the matrix. The rank of a product of two matrices is always less than or equal to the minimum of the ranks of the two matrices.
In this case, the rank of matrix $A$ is $N$. This means that there are $N$ linearly independent rows in matrix $A$. The rank of matrix $B$ is therefore at most $N$.
To determine the rank of matrix $B$, we can use Gaussian elimination. We can first eliminate the first row of matrix $B$ by subtracting $\frac{r}{p}$ times the first row from the second row. This gives us the following matrix:
$$\left[\begin{array}{cc} {p^2 + q^2} & {pr + qs} \ {0} & {\frac{r^2 + s^2 – pr – qs}{p}} \end{array}\right]$$
We can then eliminate the second row of matrix $B$ by adding $\frac{r^2 + s^2 – pr – qs}{p(p^2 + q^2)}$ times the second row to the first row. This gives us the following matrix:
$$\left[\begin{array}{cc} {p^2 + q^2} & {0} \ {0} & {\frac{r^2 + s^2 – pr – qs}{p}} \end{array}\right]$$
Since the first row of matrix $B$ is now a multiple of the second row, the rank of matrix $B$ is $N – 1$.
Therefore, the rank of matrix $B$ is $\boxed{\text{N – 1}}$.