Why is the equilibrium constant of water with its vapour equivalent to its vapour pressure?
Lets start by writing the definition of the equilibrium constant, for some general reaction:

$$\sum_i^{n_\text{reac}} a_iA_i = \sum_j^{n_\text{prod}}b_jB_j$$

The above is a general reaction, where $a_i$ and $b_j$ is the stoichiometry and $A_i$ and $B_i$ is the molecules. For this reaction we can write the equilibrium constant:

$$K=\frac{\prod_j^{n_\text{prod}}\frac{B_j^{b_j}}{U_j^{\circ}}}{\prod_i^{n_\text{reac}}\frac{A_i^{a_i}}{U_i^{\circ}}}$$

Here $\prod$ is the symbol that denotes we take the product over a range. The $U_i^\circ$ are a standard unit, to ensure that our equilibrium constant is well defined.

Now if we look at your reaction:

$$\mathrm{H_2O\left( l \right)} \rightleftharpoons \mathrm{H_2O\left( g \right)}$$

We can from the above equations identify that; $a_1 = 1$, $b_1=1$, $A_1=\left[ \mathrm{H_2O} \right]$ and $B_1=p_\mathrm{H_2O}$.

We can thus write our equilibrium constant as:

$$K=\frac{\frac{p_\mathrm{H_2O}}{1\text{ bar}}}{\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}}$$

As you might note, I have also inserted the $U$s. For pressure $U=1\text{ bar}$ and for concentrations $U=1\text{ M}$. Now we need one last ingredient. You might have learned that the activity of liquids is $1$, i.e. the means that we have to set $\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}=1$, this is due to the definition of activity. If interested I would suggest to take a look at this. As our final result we can see that:

$$K=\frac{p_\mathrm{H_2O}}{1\text{ bar}}$$

I.e. we have that the equilibrium constant equals the partial pressure of water in standard units.