I think it will be helpful to define some terms in order to understand the P versus NP problem.
P refers to all the problems that can be solved quickly or in polynomial time. In other words, exponential-time algorithms don’t count.
NP refers to decision problems that can be solved in non-deterministic polynomial time. A decision problem lends itself to being expressed as yes or no. In this context, non-determinism connotes guesses that can be made in polynomial time (Demaine, 2015). It is easy to check the correctness of solutions to NP problems even though it is difficult to find the solutions.
A problem X is NP-complete if X is in NP and X is NP-hard (Demaine, 2015). NP-complete connotes that a problem is as hard as every problem in NP. To determine that a problem is at least as hard as all problems in NP, we have to reduce all problems in NP to NP-complete. This means expressing the problem whose algorithm we don’t know in terms of a problem whose algorithm we know and then applying the known algorithm (polynomial time) to it. It should be noted that an NP-complete problem does not have a known algorithm in polynomial time.
A problem X is hard if every problem in NP reduces to X (Demaine, 2015). One can also say that a problem is hard if it is at least as hard as every problem in NP. Finding a solution to an NP-hard problem will lead to finding a polynomial time algorithm for every problem in NP (Erickson, n. d.).
P versus NP is about whether a problem than can be solved easily (in polynomial time) can also be checked easily for correctness of the solution. Nobody knows whether P = NP. If you can prove this, you will win US$1 million prize from the Clay Mathematics Institute. I think this will keep us busy for a long time.
References
Erickson, J. (n. d.). Lecture 21: NP-Hard Problems. Retrieved from http://jeffe.cs.illinois.edu/teaching/algorithms/2009/notes/21-nphard.pdf
Demaine, E. (2015). Complexity: P, NP, NP-Completeness, Reductions . Retrieved from https://www.youtube.com/watch?v=mr1FMrwi6Ew
