Converting to Chomsky normal form
Let's work it out through the conversion algorithm given in 
Wikipedia.

input:
$S \to  abC \mid  babS \mid de$
$C\to aCa \mid b$  


Introduce $S_0$:
$S_0 \to S$
$S \to  abC \mid  babS \mid de$
$C\to aCa \mid b$  
remove $\epsilon$ rules: there are no $\epsilon$ rules, so nothing changes.
eliminate unit rules: Originally, there are none, but we added one, namely $S_0\to S$. Since it's the only one, we'll deal with it later (after $S$ is already in CNF). This will be done by adding, for any rule $S\to V_iV_j$, the  rule $S_0 \to V_iV_j$. But let's first complete the conversion.
replace all other rules into normal form: we take each transition which is not in the correct form and replace it with $N\to V_iV_j$, introducing new non-terminals $V_1, V_2,...$ as needed:

$S\to ab C$ $\Longrightarrow$ $S\to V_1 C$ 
setting $V_1 \to ab$. The new $V_1$ is not in CNF, but can easily be converted to CNF by re-defining it as $V_1\to AB$ with $A\to a$, $B\to b$.  
$S\to babS$ $\Longrightarrow$ $S\to V_2S$ adding $V_2 \to bab$. Now $V_2$ is not in CNF, so we change it to $V_2 \to V_3B$ adding $V_3\to BA$. (skipping a trivial step here)  
$S\to de$ is almost CNF. We change it to $S\to DE$ and add $E\to e$, and $D\to d$.  
$C\to aCa$ $\Longrightarrow$ $C\to AV_4$ where $V_4 \to CA$.  
$C \to b$ is already in CNF.



so we end up with:
$S_0 \to S$
$S\to V_1 C$
$S\to V_2S$
$S\to DE$
$C\to AV_4$
$C \to b$
$V_1 \to AB$
$V_2 \to V_3B$
$V_3 \to BA$
$V_4 \to CA$
and $A\to a$, $B\to b$, $D\to d$, $E\to e$.

Finally, we need to deal with the unit rule $S_0 \to S$. As said, we will replace the $S$ in the right-hand-side with the "content" of S. That is, we remove that unit rule and add instead:
$S_0 \to V_1C \mid V_2S \mid DE$.