Answer
$v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$
Work Step by Step
Let $\theta$ be the angle the slope makes with the horizontal.
Then, $sin(\theta) = \frac{h}{L}$ and $cos(\theta) = \frac{\sqrt{L^2-h^2}}{L}$.
We can use a force equation to find an expression for the acceleration as the object slides down the slope;
$\sum F = ma$
$mg~sin(\theta) - F_f = ma$
$mg~sin(\theta) - mg~cos(\theta)~\mu_k = ma$
$a = g~sin(\theta) - g~cos(\theta)~\mu_k$
We can use the expression for the acceleration to find the speed at the bottom;
$v^2 = v_0^2+2aL = 0 + 2aL$
$v = \sqrt{2aL}$
$v = \sqrt{(2L)(g~sin(\theta) - g~cos(\theta)~\mu_k)}$
$v = \sqrt{(2L)(g~\frac{h}{L} - g~\mu_k~\frac{\sqrt{L^2-h^2}}{L})}$
$v = \sqrt{2gh - 2g~\mu_k~\sqrt{L^2-h^2}}$
