Does P(A|B,C)P(B)= P(A|C)? Can you remove a conditional variable by multiplying by the probability of it?
 A: We know that
$$
P(A,B)=P(A|B)P(B)
$$
and this remains true when conditional on $C$:
$$
P(A,B|C)=P(A|B,C)P(B|C).
$$
You'll notice that the $C$ conditioning seems to be "along for the ride." 
In Bayesian analysis, one typically replaces $C$ with $I$, indicating prior information. So,
$$
P(A,B|I)=P(A|B,I)P(B|I).
$$
which can get cumbersome, but it underscores the central point that every interpretation of probability is conditional on what we -- most often tacitly, but occasionally explicitly -- affirm to be true.
If you want to "remove a conditional," you may use the definition of conditional probability to restate the problem as a joint probability and a marginal:  $P(A|C)=P(A,C)/P(C).$ Where from there depends on what you are trying to do.
A: With a little probability algebra, we have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(A|B,C) 
&= \frac{\mathbb{P}(A,B,C)}{\mathbb{P}(B,C)} \\[6pt]
&= \frac{\mathbb{P}(C) \cdot \mathbb{P}(A|C) \cdot \mathbb{P}(B|A,C)}{\mathbb{P}(B,C)} \\[6pt]
&= \mathbb{P}(A|C) \times\frac{\mathbb{P}(B|A,C)}{\mathbb{P}(B|C)}. \\[6pt]
\end{aligned} \end{equation}$$
Thus, we can see that removal of the conditioning event $B$ requires us to multiply by $\mathbb{P}(B|A,C)/\mathbb{P}(B|C)$, which is not generally equal to $\mathbb{P}(B)$.
A: Not quite, as $B$ may occur or not occur on the right hand side while the left hand side is  $\mathbb P(A,B\mid C)$
But since $\mathbb P(A,B\mid C) + \mathbb P(A, B^c\mid C)= \mathbb P(A\mid C)$, you can say: 
$$\mathbb P(A\mid B,C)\mathbb P(B) + \mathbb P(A\mid B^c,C)\mathbb P(B^c)= \mathbb P(A\mid C)$$
A: It does not work out, since: 
$P(A|B,C)P(B) = \frac{P(A,B,C)}{P(B,C)}P(B) = \frac{P(A,C|B)P(B)}{P(C|B)P(B)}P(B) = P(A|C)P(B)$
For a more concrete example, take an equally probable six sided die. Let A = a 2 is rolled, B = a 4 is rolled, C = an even number is rolled. 
P(A|B,C)P(B) = 0, P(A|C) = 1/3. 
