# if P = NP, does it mean that P = NP = NP-complete?

Lets assume P = NP, so all problems in NP are decidable in polynomial time, Therefore I can solve all problems in NP in polynomial claiming P = NP = NPC.

But then, how come Σ* belongs to P = NPC because I can't reduce the even length string language (as an example of a language that is not trivial) to Σ*?

No. Even assuming $$\mathsf{P}=\mathsf{NP}$$, it is not true that all the languages therein are $$\mathsf{NP}$$-complete.
An example of a language that is in $$P$$ but it is not $$NP$$-complete (regardless of the $$P$$ vs $$NP$$ matter) is $$\Sigma^*$$, as you noticed. Indeed, there is no way to reduce any language $$L \in \mathsf{NP} \setminus \{ \Sigma^* \}$$ to $$\Sigma^*$$ since, given $$x \not\in L$$, there is no function $$f$$ such that $$f(x) \not\in \Sigma^*$$. A similar argument shows that $$\emptyset$$ is not $$\mathsf{NP}$$-complete.
However, if $$\mathsf{P}=\mathsf{NP}$$, then it is true that all languages in $$\mathsf{NP} \setminus \{\emptyset, \Sigma^*\}$$ are $$\mathsf{NP}$$-complete. To see this, let $$A \in \mathsf{NP} \setminus \{\emptyset, \Sigma^*\}$$ and pick any $$L \in \mathsf{NP}$$. Since there are $$y,z$$ such that $$y \in A$$ and $$z \not\in A$$, a valid Karp reduction from $$L$$ to $$A$$ is the following: $$f(x) = \begin{cases} y & \text{if } x \in L \\ z & \text{if } x \not\in L \end{cases}.$$ Notice that $$f$$ can be computed in polynomial time since $$L \in \mathsf{NP} = \mathsf{P}$$ by hypothesis (and hence it is possible to check whether $$x \in L$$ in polynomial time).