Showing that $\mathbb{E}_c[p(a \mid b, c)] = p(a \mid b).$ It's simple to show that (note that all the expectations are wrt the marginal distribution of corresponding variable, e.g. $\mathbb{E}_b[f(b)] = \int f(b) p(b) \text{ d}b = \int f(b) (\int p(a, b, c) \text{ d} a \text{ d} c) \text{ d}b$)
$$
\mathbb{E}_b[p(a \mid b)] = p(a),
$$
simply by evaluation as
$$
\mathbb{E}_b[p(a \mid b)] =  \int p(b) p(a 
 \mid b) \text{ d} b  = \int p(b) \frac{p(a, b)}{p(b)} \text{ d} b = p(a). 
$$
But I'm struggling in showing a conditional version of this in the form of
$$
\mathbb{E}_c[p(a \mid b, c)] = p(a \mid b).
$$
How to show the last relation? Surely it must hold since it is just a conditional version of the first equation.
 A: As stated in the answer below, the equality you want to show holds if $c$ is independent of $b$. With your formal notation:
\begin{align}
p(a\mid b) &= \frac{p(a,b)}{p(b)}\\
&=\int \frac{p(a,b,c)}{p(b)} dc\\
&=\int \frac{p(a \mid b, c)p(b, c)}{p(b)} dc\\
&=\int \frac{p(a \mid b, c)p(c \mid b)p(b)}{p(b)} dc\\
&=\int p(a \mid b, c)p(c \mid b) dc \\
&= \mathbb{E}_{c\mid b}[p(a\mid b,c)]
\end{align}
If $c$ is independent of $b$, then $p(c\mid b) = p(c)$ and you have what you want.
Edit: see here for the discrete case
A: According to the comments of the OP, I can just guess the question. So, let's assume $A$, $B$ and $C$ are random variables with densities on $\mathbb R$. Then the conditional density of $A$ given $B$ and $C$ is defined as
$$f_{A|B=b,C=c}(a):=\frac{f_{A,B,C}(a,b,c)}{f_{B,C}(b,c)}$$
where the right hand side is the ratio of the known joint densities.
Then the questions is, what happens if we integrate over the values of $C$ and its density:
$$\int_{\mathbb R}f_C(c)f_{A|B=b,C=c}(a)\,dc=\int_{\mathbb R}f_C(c)\frac{f_{A,B,C}(a,b,c)}{f_{B,C}(b,c)}\,dc=f_{A|B=b}(a)$$ only if $B$ and $C$ are independent, i.e. $f_{B,C}(b,c)=f_B(b)f_C(c)$.
