Why does increasing the available surface area of a pure solid not effect equilibrium?
You can change the amount of $\ce{B}$ available and the forward rate, but that doesn't change the rate constant. And the equilibrium constant is directly related to the rate constants of the forward and reverse reactions:

$$K = \frac{k}{k'}, \text{where }k\text{ is forward rate cosntant, }k'\text{ is reverse rate constant}$$

As @MaxW points out, you'll get to equilibrium faster, but it's the same equilibrium.

Edit:

Here's another way to think about it. Consider the reaction:
$$\ce{H2O_{(l)}&lt;=&gt;H2O_{(g)}}$$

Does the equilibrium between liquid water and water vapor change if you increase the surface area of a pool of water? No, because even though evaporation is faster, there's also more area for condensation to take place.

In your specific case, the reverse reaction from $\ce{C}$ needs to interact with some amount of solid to redeposit $\ce{B}$. Otherwise, you're making $\ce{B_{(g)}}$ and that's not the same thing...

Edit 2 (for OP's comments):
Notice that your reaction is actually extremely unrealistic. Gases don't just form solid or liquids on their own; they need some nucleation seed. Otherwise, there would be no point in seeding rain clouds.

As @IvanNeretin points out, invoking microscopic reversibility, you'd need to invoke these nucleation points in your reaction, forward and backward. These points are of course whatever surface is available on $\ce{B}$.

This makes the reaction:
$$\ce{A_{(g)} + B_{(g)} + n B_{*} \overset{k}{\underset{k'}{&lt;=&gt;}} n B_{*} + C_{(g)}}$$

$\ce{B_{*}}$ refers to the surface. Notice that it cancels out so it doesn't affect the equilibrium constant, but having more surface area will increase the forward rate if you increase the surface. It will also increase the reverse rate, which is why the equilibrium constant can stay the same.