- Problem A is in NP if:
B is in NP and you can reduce B to A in polynomial time -> B $\leq_p$ A
Not quite right. The class P is a subset of NP anyway, and hence A is already NP if it is in P. The question is whether A is in P or not.
By reducing B to A in polynomial time, you prove that any polynomial solution of A is a polynomial solution of B. Assuming B is in NP gives no useful information about A, since it is possible that B is in P as well (remember that P $\subseteq$ NP). However, assuming (and probably that is what you meant), that B is NP-Hard, then, by reducing B to A, we prove that A is also hard for NP class, and hence unless P=NP, the problem A is not in P.
Now back to the question. What we know so far, P $\subseteq$ NP. There are more classes above NP like EXP and under P like L for example. Which means,
$$L \subseteq P \subseteq \mathit{NP} \subseteq \mathit{EXP}.$$
Check the link for more information about the classes. However, one characterization of problems in NP, is that they all admit a certifier, that can in polynomial time in the size of the input, say if the given instance is a yes instance. For example, for travelling salesman, which is an NP-Complete problem, the certificate is a path. You can clearly certify it, by checking if it visits all cities and each city exactly once and return to the beginning. Since P $\subseteq$ NP the same works for P problems. Like shortest path. given a shortest path between two vertices, you can certify it it is a shortest path. For example, the certifier can compute a shortest path in polynomial time and compare the lengths.
Now the interesting part about certificates, is to certify if a problem admits a certificate for the negative answer. That means, given an instance and a certificate, the certifier can tell in polynomial time, if the instance is not in the language. Problems that admits such a certifier belong to the class co-NP. It is believed that NP $\neq$ co-NP unless P = NP. The class P belongs to the intersection of both classes and hence we can certify both negative and positive answers in polynomial time.
Summary
Reducing $A$ to a polynomial problem proves that it is solvable in polynomial time.
Reducing an NP-Hard problem to $A$, proves that $A$ is NP-Hard and probably not in P. In this case, even if the problem is in NP and hence is NP-complete, we can only say about $A$ that it is NP-hard, we still do not not if it is in NP.
A problem is in NP if there is for each instance that belongs to language of the problem, a certificate, that can be certified in polynomial time in the size of the instance.
A problem is in co-NP if there is for each instance, that odes not belong to the language of the problem, a certificate, that can be certified in polynomial time in the size of the instance.
The class P belongs to the intersection of P $\cap$ co-NP.
