This may be a naive question, but I'll pose it.

Is there an example of a notion of forcing $\mathbb{P}$ that has the $\kappa$-c.c. which is *not* also $\kappa$-Knaster Property also is not "factorable" as a product of partial orderings?

For reference, let me state the relevant definitions and a typical example.

Definition. A partial order $\mathbb{P}$ is $\theta$-c.c. if there is no antichain $A\subseteq\mathbb{P}$ of size $\kappa$.

Definition. A partial order is $\kappa$-Knaster if, for every sequence $\overline{p}=\langle p_\alpha\rangle_{\alpha<\kappa}$ of elements of $\mathbb{P}$ of size $\kappa$, there is an unbounded set $X\subset\kappa$ of ordinals such that $\langle p_\beta\rangle_{\beta\in X}\subset \overline{p}$ which is pairwise compatible.

The prototype example involves *a product* of Souslin trees:

If $T$ is a Souslin tree, then $T$ has $\omega_1$-c.c. and is therefore also also $\omega_1$-Knaster. However, $T\times T$ is not $\omega_1$-c.c., yet this *product* is $\omega_1$-Knaster.

Is this an accident of the separability of $T$ or are there "natural" (perhaps even non-separative) examples of partial orders that are not "resolvable" or "decomposable" into products (Cartesian or otherwise) of other partial orders?

Edit:

After going through much confusion and re-reading my source text several times, I realize that the "prototypical example" I gave above is *not* $\omega_1$-Knaster. In fact, the very text I was reading from explicitly states this and for whatever reason, I did not see the word "not" there. It turns out this is the most important word in that particular sentence, so many thanks are due to Paul McKenney for raising the issue.

I think my original question was intended to be whether the Knaster property can hold for partial orders that are not decomposable into products of some sort, but obviously I got carried away trying to give some background.