If the Clique-k Problem is in P, why not Clique as well?
It's a matter of logical quantifiers. 

Consider this statement: "if every natural number $n$ is bounded above by a constant $c$, how come there is no constant $c$ that bounds every natural number $n$ by above?".

The above is an example where $\forall n.\ \exists c.\ \ldots$ does not imply
$\exists c.\ \forall n.\ \ldots$. Indeed, in the former, we can choose $c$ after we see $n$, while in the latter we must do before, and we are unable to do that: whatever $c$ we choose, the statement becomes false for e.g. $n=c+1$.

To decide CLIQUE, once we take the instance $(G,k)$ to be checked,
we can use an algorithm which runs on $O(n^c)$ for some $c$ -- but here we are choosing $c$ after we have seen $k$. This does not prove that CLIQUE is in P. For that, we would need to find a constant $c$, and an associated $O(n^c)$ algorithm, which would decide any instance $(G,k)$, whatever $k$ is. And this is much harder.