Building a sequence that approximates given sequences Suppose that we are given three sequences $a1,a2$ and $a3$  each describing a total ordering on $N$ 'entities'. For example,
$$\langle a1\rangle=1<9<8<2<3<\cdots<N $$
means that entity #1 is worst and entity #N is the best according to $a1$. 
I need to come up with a new total ordering, let us call it $\langle x \rangle$ that best approximates all given orderings.
One way I was thinking of doing this task with $N=4$:
$$<a1> = 1<3<4<2 $$ $$<a2> = 1<2<4<3 $$
 $$<a3> = 4<2<3<1 $$
Look at each pair $x,y \in \lbrace 1,2,3,4 \rbrace, x\neq y $. For example let us look at the tuple $(1,4)$. Clearly the first two sequences claim $1<4$ while the third claims $4<1$. Pick out the majority winner. Here, $1<4$ wins.
Do this for each possible tuple. We then have $\binom{N}{2}$ such tuples. 
List them , for eg: $1<4$, $2<3$, ...
Use these now to combine them into a total ordering $<x>$. 
ISSUE: This problem can be converted to the Hamilton Path problem. Let each number be a vertex. Put a directed edge between $x$ and $y$ iff $x<y$. Now find the directed path that convers each vertex exactly once, this is my $<x>$.
Is my argument that this problem is NP complete correct? (Since Ham path is NP complete). If so, in what other ways can I do this task? 
 A: 
Is my argument that this problem is NP complete correct? (Since Ham path is NP complete).

The argument is both incorrect and wrong.
It's incorrect because reduction arguments have to go the other way: by solving your problem you can solve any instance of a known NP-complete problem (and the reduction is polynomial).
It's wrong because the graph defined in your problem is a tournament (Wikipedia, MathWorld) and a Hamiltonian path in a tournament on $n$ vertices can be found in $O(n \lg n)$ time. See e.g. 
Bang-Jensen and Gutin, On the complexity of hamiltonian path and cycle problems in certain classes of digraphs, for a survey.

Although it's not directly what you ask, I should note a problem with the proposed algorithm:

Now find the directed path that convers each vertex exactly once, this is my $<x>$.

(my emphasis). There isn't necessarily a unique Hamiltonian path. Consider the minimal example where the three input orders are $$1 < 2 < 3 \\ 2 < 3 < 1 \\ 3 < 1 < 2$$
The graph is a cycle $1 \to 2 \to 3 \to 1$, so there are three Hamiltonian paths. Obviously in this case the three are equally good / bad approximations, but maybe a more complicated example which embeds them would have Hamiltonian paths which vary in how many of the original pairwise comparisons they contradict.
