$L = \{cda^nb^n\mid n\in \Bbb N\} \cup \{a,b,d\}^*$
Assuming $L$ is regular then there exist a pumping length $n$ for $L$. Lets use w = $cda^nb^n$.
Thus $w \in L$ and $|w| = 2n+2$ $\implies$ $|w| \geq n$.
$w$ can be splitted into three pieces $w = xyz$ with the following conditions:
- $|xy| \leq n$
- $|y| \geq 1$
Let be $x = cda^j$ and $y = a^k$ with $ k+j\leq n$ and $ k\geq 1$.
Now choose $i = 2$ so that $xy^iz = xyyz = cda^ja^ka^ka^{n-j-k}b^n = cda^{n+k}b^n$
$w$ has now more $a$'s than $b$'s (considering that $k$ is at least 1) $\implies w \notin L \implies$ $L$ is not regular.
Is that enough for the proof?