Problem: Assume $\Pi:\mathbb N^2\to\mathbb N\setminus\{0\}$ is a primitive recursive coding of the pairs of numbers, that is also a bijection and $(\forall (x,y)\in\mathbb N^2)(\Pi(x,y)>max(x,y))$.
Let $\mathbb N^*$ be the words on the alphabet of natural numbers and $\epsilon$ be the empty word. Let
$$k:\mathbb N^*\to\mathbb N:$$ $$k(\epsilon)=0$$ $$k(\langle x_1\dots x_n\rangle)=\Pi(x_1,k(\langle x_2\dots x_n\rangle)$$
Show that $k$ is an effective coding of $\mathbb N^*$
Some thoughts: I need to show that $k$ is a bijection and that its length and member functions are primitive recursive. I can show that $k$ is injective on my own. How can I show that $k$ is surjective? I tried using course-by-values induction on the length of the word but failed. Additionally, how can I show that the length function is primitive recursive? Having the above, I can show that the member functions are primitive recursive.
p.s. This is part of the problem only. I am working with a given $\Pi$, that I have show to be a primitive recursive coding. Moreover, using that $k$ is an effective coding I have shown that a bunch of other functions are primitive recursive. To avoid a long post, however, I have left out those details.
The assumptions in the question is not enough for your $k$ to work.
For example if $\Pi(x,y)=\frac{(x+y)(x+y+1)}2+x$ (the Cantor zig-zag, a fairly standard choice of pairing function), then $\Pi(0,0)=0$ and hence $$ k(\langle \rangle) = 0 = k(\langle 0 \rangle) = k(\langle0,0\rangle) = \cdots$$
You need the additional assumption that the range of $\Pi$ is exactly $\mathbb N\setminus\{0\}$ (as is the case for the $\Pi$ you give in the comments) -- then $k$ will a bijection.