How to figure out the state of aggregation of an unknown substance via properties of a chemical equilibrium in a closed system? This is a question from the College Entrance Test in China.

In a closed, constant temperature container, reaction $\ce{A(s)&lt;=&gt;B +C(g)}$ reached chemical equilibrium. If the volume of the container is reduced, when the equilibrium is reached again, the concentration of $\ce{C}$ equals the concentration of $\ce{C}$ before the reduction.
Is it possible that $\ce{B}$ is in gas state?

The official answer is impossible.
And the official explanation to this is:

If $\ce{B}$ is in gas state, we have $K=C_\ce{B} \cdot C_\ce{C}$
Otherwise, if $\ce{B}$ is in liquid or solid state, we have $K=C_\ce{C}$
Since the concentration of $\ce{C}$ does not change after the reduction, in order to remain the $K$ value unchanged, $C_\ce{B}$ must not be involved in this process. So $\ce{B}$ is impossible to be in gas state.

But I don't think it makes sense.
Consider a container with a volume of $V$, $b$ mol of $\ce{B(g)}$ and $c$ mol of $\ce{C(g)}$ is inside it.
If   $K_{\text{pre}}=K_{\text{post}}$
, we have $C_{\ce{B},\text{pre}} \cdot C_{\ce{C},\text{pre}}=C_{\ce{B},\text{post}} \cdot C_{\ce{C},\text{post}}$.
When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{\text{post}}}{b_{\text{pre}}}=\dfrac{c_{\text{post}}}{c_{\text{pre}}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.
Conclusion: $\ce{B}$ can be in gas state.
So, is the official answer wrong or I just didn't make the right conclusion?
Distinction: you are only considering Kp, which is an equilibrium constant that doesn't take concentration into account. You're going wrong here because the question asks for concentration, which has a unit of $L^{-1}$, not mole fraction, which is just in units of moles. 
$$Kc=[C]c[D]d / [A]a[B]b$$
$$Kp=(C)c(D)d / (A)a(B)b$$
Think about their answer to the question. It states that $K=Cc_{pre}$ and $K=Cc_{post}$. The only way for this to make sense is if $Cc$ is in units of concentration because then as the total moles of c go up, the fact that the volume is decreasing doesn't change the value of Kp.

When you say,


  If   $K_{pre}=K_{post}$ 
  , we have $C_{\ce{B},pre} \cdot C_{\ce{C},pre}=C_{\ce{B},post} \cdot C_{\ce{C},post}$
  
  When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{post}}
{b_{pre}}=\dfrac{c_{post}}{c_{pre}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.


you do not take concentration into account, or the fact that $Cc_{pre} = Cc_{post}$

For then you cannot restate your equation in terms of concentration: $\dfrac{Cc_{post}}{Cc_{pre}}=\dfrac{1}{2}$