How to calculate Ka for hydronium and Kb for hydroxide?
To an extent, you are right; the calculation of the $K_\mathrm{a}(\ce{H2O})$ does include $[\ce{H2O}]$.

Consider the following equation:
$$\ce{H2O(l) +H2O(l) -&gt;H3O+(aq) + OH- (aq)}$$

The equilibrium for this equation is:

$$K_\mathrm{a}(\ce{H2O})=\frac{[\ce{H3O+}]\cdot [\ce{OH-}]}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Since $K_\mathrm{w}=[\ce{H3O+}]\cdot [\ce{OH-}]=1.01 \times 10^{-14}$ at $25^\circ \mathrm{C}^{[1]}$, we can make such a substitution in our equation:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]\cdot [\ce{H2O}]}$$

Now here you will have to bear with me. In a usual $K_\mathrm{a}$ calculation, $[\ce{H2O}]$ is disregarded, as it remains mostly constant ($55.5084\ \mathrm{M}$), except when working with extremely concentrated solutions. In this equation, there are two $[\ce{H2O}]$ terms, and only one of them is disregarded (as a solvent). The other one can be considered concentration of the acid. Therefore:

$$K_\mathrm{a}(\ce{H2O})=\frac{1.01 \times 10^{-14}}{[\ce{H2O}]}=\frac{1.01 \times 10^{-14}}{55.5084}=1.819544 \times 10^{-16}$$

Thus we have calculated the $K_\mathrm{a}$ for water, or the $K_\mathrm{b}$ for $\ce{H3O+}$. The relation between these is:

$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$

So:

$$K_\mathrm{a}(\ce{H3O+})=\frac{1.01 \times 10^{-14}}{1.819544 \times 10^{-16}}=55.5084$$

and:
$$-\log(K_\mathrm{a}(\ce{H3O+}))=\mathrm{p}K_\mathrm{a}(\ce{H3O+})=-\log(55.5084)=-1.7404$$

The calculation for the $K_\mathrm{b}(\ce{OH-})$ is exactly the same for the calculation of $K_\mathrm{a}(\ce{H3O+})$ (pretty much just replace all the $\ce{H3O+}$ with $\ce{OH-}$) and yields the same result.



[1] Starkey, R.; Norman, J.; Hintze, M. Who Knows the Ka Values of Water and the Hydronium Ion? J. Chem. Educ. Journal of Chemical Education. 1986, 63, 473.