How to marginalise this? I came across this example in one of my lectures, but am unable to figure out how to carry out the marginalisation over c. In the context of bayesian networks:

For the first expression, it is obtained by reading the conditional independence assertations from the graph:
$p(ABC) = p(a)p(c|a)p(b|c)$
From the LHS of the first expression we can get:
$\Sigma_cp(ABC) = p(AB)$
But for the RHS of the first expression how to we get from:
$\Sigma_c(p(A)p(C|A)p(B|C)) = p(A)p(B|A)$

The marginalization is just another form of applying the law of total probability:
\begin{align}
 & p(a, b) \\
=& p(a)p(b|a) \tag{1} \\
=& p(a)\sum_c p(b, c|a) \tag{2} \\
=& p(a)\sum_c p(b|c, a)p(c|a). \tag{3} \\
=& p(a)\sum_c p(b|c)p(c|a). \tag{4}
\end{align}
This completes the proof. Note that $(1)$ -- $(3)$ hold for three arbitrary random variables, whereas the correctness of $(4)$ must rely on the given graphical model specification.

Explanations:
$(1)$:  Using the definition of conditional probability.
$(2)$:  The event $\{B = b\}$ can be expressed as $\{B = b\} = \bigcup_c (\{B = b\}\cap\{C = c\})$.  This is essentially the same as deriving the law of total probability.
$(3)$:  Apply the definition of conditional probability again to verify $p(b, c|a) = p(b|c, a)p(c|a)$.
$(4)$:  For the given the graphical model, $A$ and $B$ are independent conditional on $C$, whence $p(b|c, a) = p(b|c)$.
