Equilibrium constant for a multi-step reaction
To give a specific example, let's take this set of reactions:

$$\tag{1}\ce{A + B &lt;=&gt; I_1}$$

$$\tag{2}\ce{I_1 + C &lt;=&gt; I_2}$$

$$\tag{3}\ce{I_2 &lt;=&gt; I_3 + D}$$

$$\tag{4}\ce{I_3 &lt;=&gt; E + F}$$

To get the net reaction, we add up the equations and cancel the intermediates:

$$\ce{A + B + C &lt;=&gt; D + E + F}$$

When the system is a equilibrium, all the reactions are at equilibrium. You can write the equilibrium constant expression based on the net reaction:

$$K_\text{net} = \frac{[\ce{D}][\ce{E}][\ce{F}]}{[\ce{A}][\ce{B}][\ce{C}]}$$

You would arrive at the same expression if you take the product of the individual equilibrium constants (and cancel):

$$K_\text{net} = K_1 K_2 K_3 K_4 =\frac{[\ce{I_1}]}{[\ce{A}][\ce{B}]} \times \frac{[\ce{I_2}]}{[\ce{I_1}][\ce{C}]} \times \frac{[\ce{I_3}][\ce{D}]}{[\ce{I_2}]} \times \frac{[\ce{E}][\ce{F}]}{[\ce{I_3}]}$$

The concentrations of the intermediates will not be zero because they are in equilibrium with the other species. They are intermediates in the sense that we care a bit less about them since they don't appear in the net equation of interest. However, they are part of the mass balance and may well be among the major species.