The answer depends on what you mean by the notation $\le_p$.

If $\le_p$ refers to a [Karp reduction](https://en.wikipedia.org/wiki/Polynomial-time_reduction) (i.e., a many-one reduction), then the answer is Yes.  It follows from the definition of NP.  See http://cs.stackexchange.com/q/9556/755 to learn more.

Assuming that $\le_p$ refers to a [Cook reduction](https://en.wikipedia.org/wiki/Turing_reduction) (also known as a polynomial-time Turing reduction), then the answer is "We don't know, but we think not".  Let $X$ be the problem TAUTOLOGY, and $Y$ be the problem SAT.  Then $X \le_p Y$ (we can transform an instance of TAUTOLOGY to an instance of SAT; if you then flip the answer from an algorithm to solve SAT, the result will be the answer to TAUTOLOGY).  We know that SAT is in NP.  However $X$ is [co-NP-complete](https://en.wikipedia.org/wiki/Co-NP-complete), so $X$ is not likely to be in NP: if $X$ is in NP, then it follows that NP = co-NP, which would be a surprise.  While we can't prove NP $\ne$ co-NP, many/most complexity theorists expect that NP $\ne$ co-NP is true, and from that it would follow that the answer to your question is "No" -- assuming you are referring to Cook reductions.

To learn more about the difference between these two types of reductions and why both notions exist, take a look at the following questions:

- http://cs.stackexchange.com/q/9847/755

- http://cs.stackexchange.com/q/11120/755

- http://cstheory.stackexchange.com/q/138/5038

- http://cstheory.stackexchange.com/q/686/5038