Let $D$ be a distribution. Show that every point of $p \in M$ has a submanifold $N$ such that $p \in M$ and $T_p N = D_p$. Let $D$ be a distribution. Show that every point of $p \in M$ has an embedded submanifold $N$ such that $p \in M$ and $T_p N = D_p$.

Attempt: Assume $M$ is a smooth manifold of dimension $m$ and $D$ is a rank $k$-subbundle of $\pi: TM \to M$. 

Let $p \in M$. Somehow, we will have to construct an embedded submanifold $N$ of dimension $k$ such that $D_p = T_p N$. 

How can I construct such an embedded manifold? Any hints will be appreciated!
The problem is to find a submanifold that takes a certain tangent space at a certain point. You can do this when your manifold is an open subset of $\mathbb{R}^n$ trivially; just take a little piece of an affine space. Luckily, every manifold is locally diffeomorphic to an open subset of $\mathbb{R}^n$.

More specifically: let $\varphi\colon U \to \mathbb{R}^n$ be a chart of the manifold at $p$. Let $N$ be an embedded submanifold in $\varphi(U) \subseteq \mathbb{R}^n$ that takes the tangent space $(d \varphi)_p(D_p)$ at $\varphi(p)$. (For example, let $N$ be a restriction of an affine subspace.) Then, $\varphi^{-1}(N)$ is an embedded submanifold of $U$, and hence of $M$, that takes the tangent space $D_p$ at $p$. (To verify this, it is most convenient to regard an embedded submanifold as an inclusion map, and the tangent space "taken at a point" as the image of the differential of this inclusion.)
This is not always true there exist  distributions which are not integrable.

https://mathoverflow.net/questions/294276/a-strongly-non-integrable-distribution