My attempt(almost sure that it is not quite working):
It is known that the greatest cardinal is P (ω) which is not a well-orderable set. Another fact is that P (ω) is a regular cardinal which by the definition give us :every unbounded subset A<P (ω) has the same cardinality as P (ω). But by the property of cardinals, namely that if two sets P (ω) and A have the same cardinal numbers, then they are equinumerous,the problem is solved.However, I have just used some well known propertis. Is there any way to complete the proof using that:
A chain in <P (ω), < > will be such a subset A< P(ω)that for every two elements in it they will be (<=) comparable. With other words, A<P (ω) is a chain, if <A,P (ω)∩(AxA)> is linear ordered.
None of what you've written works.
First of all, none of the three facts you claim at the outset (that $\mathcal{P}(\omega)$ has the greatest cardinality, that it is not well-orderable, and that its cardinality is regular) is justified. The first is provably false: for the first consider e.g. $\mathcal{P}(\mathcal{P}(\omega))$ and more generally note that there is no largest cardinality. The second is independent of $\mathsf{ZF}$, and in particular it is consistent with $\mathsf{ZF}$ that every set can be well-ordered (this is exactly the axiom of choice). Finally, the third is even undecidable from $\mathsf{ZFC}$ axioms: it is consistent, for example, that $\vert\mathcal{P}(\omega)\vert=\aleph_{\omega_1}$, which is singular.
Moreover, even if we were to grant those assumptions your subsequent claim that every unbounded set in $(\mathcal{P}(\omega), \subseteq)$ has size continuum would also unjustified. Indeed, that claim is false: consider e.g. $\{\omega\}$, which is a one-element unbounded subset of $\mathcal{P}(\omega)$.
The actual solution to the problem, incidentally, is constructive: we can whip up an explicit example of a size-continuum chain in $\mathcal{P}(\omega)$. This problem has appeared repeatedly on this site. One hint I have is to replace $\omega$ by $\mathbb{Q}$ (since they have the same cardinality, the two powerset-posets will be isomorphic) and think about a particular kind of set of rational numbers which you've already seen in real analysis ...