How to figure out the state of aggregation of an unknown substance via properties of a chemical equilibrium in a closed system?
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: $\dfrac{Cc_{post}}{Cc_{pre}}=\dfrac{1}{2}$