If the equation $\mathrm{k} x y+10 x+8 \mathrm{y}+16=0$ represents a pair of lines, then

The condition for a second-degree equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ to represent a pair of lines is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.

Comparing $kxy + 10x + 8y + 16 = 0$ with the general form yields:

$a = 0$, $b = 0$, $h = \frac{k}{2}$, $g = 5$, $f = 4$, $c = 16$

Substituting into the condition:

$(0)(0)(16) + 2(4)(5)\left(\frac{k}{2}\right) - (0)(4^2) - (0)(5^2) - 16\left(\frac{k}{2}\right)^2 = 0$

$20k - 4k^2 = 0$
$4k(5 - k) = 0$

The solutions are $k = 0$ and $k = 5$, both representing degenerate pairs of lines.

The final answer is $\boxed{C}$