How am I to interpret the phrase "a solution of acid"?
The acidity constant is defined as the equilibrium constant for the reaction:

$$\ce{HA &lt;=&gt; H^+ + A^-}$$

$$K_\text{a} = \dfrac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$

Note: $\ce{H3O+}$ can be considered equivalent to $\ce{H+}$

Since you have to find an equilibrium constant, (and no further info is given) you can safely assume the solution to be in equilibrium.

Plug in the various conductivities in the second equation (Kohlrausch's Law) you mentioned getting the limiting conductivity ($\lambda^o$) of the acid. This is the conductivity of the acid if it had completely dissociated into its constituent ions.



EDIT: To find the limiting conductivity of the acid using the equation, you must have the values of $\lambda^o_{\ce{H^+}}$ and $\lambda^o_{\ce{A^-}}$.

$$\lambda^o_{\ce{HA}} = \lambda^o_{\ce{H+}} + \lambda^o_{\ce{A-}}$$



The degree of dissociation of the acid $\alpha$ (The number of acid molecules which have dissociated at equilibrium per each acid molecule present originally -- it is fractional for weak acids and bases) would be the ratio of the measured conductivity of the solution ($\lambda$, which is given) to the limiting conductivity ($\lambda^o$).

$$\alpha = \dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}}$$

This alpha would be the same as the ratio of concentration of dissociated acid molecules at equilibrium (since each dissociated acid molecule gives one $\ce{H^+}$ ion, this is equal to the number of $\ce{H^+}$ ions at equilibrium) to the initial concentration of acid taken (which you have mentioned as given).

Therefore,
$$\dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}} = \dfrac{[\ce{H^+}]}{[\ce{HA}]_\text{initial}}$$

You can find $[\ce{H^+}]$ at equilibrium using this equation and then find the other concentrations using that. 

Note that all this assumes that the acid is monoprotic (i.e. there is only one proton released per acid molecule). You would have to change all the equations accordingly in case more than one proton is released.