Degree of association, van't Hoff factor and dissociation constant
There is absolutely no difference in your two equations but you forgot the units which makes it seem like there is a difference.

Indeed it is in both cases $$k_a=\frac{c~\alpha^2}{1-\alpha}$$
since what happens if your concentration would be $c=1~\mathrm{mol~L^{-1}}\approx1~\color{\red}{\text{forgotten unit}}$?


   $$k_a=\frac{\alpha^2}{1-\alpha}~\ce{mol~L^{-1}}$$




How do I know that the c is the concentration?

Using the acid dissociation constant $$k_a=\frac{c(\ce{A-})~c(\ce{H3O+})}{c(\ce{HA})}$$ we have a good basis to obtain what we search. Now by including the degree of dissociation we can define equations for the undissolved acid molecules $c(\ce{HA})$ and the concentration of the conjugate base in solution $c(\ce{A-})$ $$\begin{align}c(\ce{HA})=&amp;~(1-\alpha)~c_0(\ce{HA})\\c(\ce{H3O+})=c(\ce{A-})=&amp;~\alpha~c_0(\ce{HA})\end{align}$$ and combine it with the equation for the acid dissociation constant to obtain $$k_a=\frac{(\alpha~c_0(\ce{HA}))^2}{(1-\alpha)~c_0(\ce{HA})}=\frac{c_0(\ce{HA})~\alpha^2}{1-\alpha}$$
and this is exactly your equation which tells us that the $c$ is indeed a concentration ... namely the concentration of your solution.