Proving one time pad is perfectly secret
First, be careful about case:


The uppercase letters denote random variables in some probability space.
The lowercase letters denote particular values in the domains of the corresponding random variables.


For example, if the probability space is $\Omega$, then $C\colon \Omega \to \{0,1\}^\ell$ is a random variable for the ciphertext in the one-time pad system, and $c \in \{0,1\}^\ell$ is a particular value of a ciphertext, which at some sample $\omega \in \Omega$ might be the value of $C(\omega) = c$.

The random variable $C$ is related to the random variables $M$ and $K$ by $C = M \oplus K$, which is the standard mild abuse of notation for random variables meaning that $C(\omega) = M(\omega) \oplus K(\omega)$ in any sample $\omega \in \Omega$.  Any fixed values $c$, $m$, and $k$ might or might not be related by $c = m \oplus k$, although if in some sample $\omega$ we have $C(\omega) = c$ and $M(\omega) = m$ and $K(\omega) = k$ then $c$, $m$, and $k$ are indeed related by $c = m \oplus k$.

For any fixed message $m \in \{0,1\}^\ell$, in the event $E = \{\omega \in \Omega \mathrel: M(\omega) = m\}$ where the message $M$ takes on the value $m \in \{0,1\}^\ell$, what can you conclude about $C$?  In other words, if for any $\omega$ we have $M(\omega) = m$, what can you say about $C(\omega)$?  Can you use this to eliminate $C$ from the ellipsis in $\Pr[\;\cdots \mathrel| M = m]$?

Next, if you have an equation about the random variable $K$ without reference to $M$, given an equation about $M$, recall the definition of independence of random variables and how it relates to conditional probabilities.  Finally, use the probability mass function for $K$, and you'll have an answer.