Possible to have $m$-dimensional $C^k$ embedded submanifold in $\mathbb{R}^p$ with canonical projections onto $m$ dimensions are of dim $Notations/terminologies: 
Let $\lambda_m$ be the Lebesgue measure on $\mathbb{R}^m$.
Let $\sigma \in \Sigma_p$ be a permutation of the symbols $\{1,2...p\}$.
Let $k\mathbb{N}$ denote a desired degree of smoothness of the manifold in question.
Let $M^m \subset \mathbb{R}^p$ be a $C^k$ embedded submanifold of dimension $m$. Does this always mean that: at least one of the canonical projections onto a collection of  $m$ canonical (Euclidean) co-ordinates give rise to/have an image $N$, so that $N \subset \mathbb{R}^m$ is also $\lambda_m$ almost everywhere a submanifold (with or without boundary) of $\mathbb{R}^p$ of dimension exactly $m$? I.e. is it true that, for any canonical projection $\pi: (x_1,...x_p) \mapsto (x_{\sigma(1)},x_{\sigma(2)},...x_{\sigma(m)})$, where $\sigma \in \Sigma_p, N:=\pi(M)$ is also a $\lambda_m$ almost everywhere a submanifold?
In other words, my question is: is it possible to construct a submanifold $M^m \subset \mathbb{R}^p$ so that all of the ${p \choose m}$ canonical projections onto any of the $m$ co-ordinates have dimension strictly less than $m$ for a subset of positive $\lambda_m$ measure?
Attemp I: The way I was thinking of constructing a counter example is by constructing a non-linear map from a low dimension to a high dimension so that there's enough co-ordinates to make the map invertible, but none of the low dimensional projections are invertible. E.g. I was thinking of constructing a map:
$F:\mathbb{R}^2\to \mathbb{R}^3$ given by: $F(x,y):=(xy^2, x^2y, xy + x^2y)$ so that we can check that projection onto any of the two co-ordinates have manifold dimension $1$. But I'm not sure if it's possible to construct such a counterexample?
Attemp II: I think for any smooth submanifold in question defined using global charts (like the ones I was trying in Attempt I) wouldn't give us a counter example. This is because, locally, any smooth submanifold of dimension $m$ can be written as $F(x_1,...x_m)\mapsto (y_1,...y_p)$ where $y_i = y_i(x_1,...x_m)$ is a smooth function so that the Jacobian $[\frac{\partial y_i}{\partial x_j}]_{1 \leq i \le m, 1 \leq j \leq p }$ is of rank $m$, hence an $m \times m$ submatrix, which WLOG can be assumed to be $[\frac{\partial y_i}{\partial x_j}]_{1 \leq i \le m, 1 \leq j \leq m }$ is of full rank $m$. But then the projection $(y_1,...y_m, y_{m+1},...y_p) \mapsto (y_1,...y_m)$ gives us locally a manifold of dimension $m$. But I guess we can't construct a global such projection?
 A: I will assume that $M$ is nonempty (by the definition, the empty set is a manifold of every dimension $m$). I also assume that you want a projection $\pi$ to some coordinate subspace $R^m$ of dimension $m$ such that $\pi(M)$ has nonempty interior and topological boundary (frontier) of zero measure. 
Such projection indeed exists. First, some simple linear algebra fact: if $V$ is an $m$-dimensional linear subspace of $R^p$ then there exists a coordinate subspace $W\cong R^m$ in $R^p$ such that the natural projection $\pi: R^p\to W$ satisfies $\pi(W)=V$. This is an immediate consequence of the fact that the row-rank of a matrix equals its column-rank. 
Pick a point $p\in M$. The tangent space $T_pM$ is a $p$-dimensional linear subspace $V$ in $R^n$. Hence, as noted above, there is a coordinate subspace $W=R^m\subset R^n$ such that the restriction to $V$ of the natural 
projection $\pi: R^n\to W$ is surjective, i.e. has rank $m$. The inverse mapping theorem (or the implicit function theorem if you prefer) then implies that there exists a neighborhood $U$ of $p$ in $M$ such that the restriction of $\pi$ to $U$ is a diffeomorphism to its image and the image is an open subset of $R^m$. Let $S\subset M$ denote the subset consisting of points $q$ such that $d\pi: T_qM\to W$ has rank $<m$. By Sard's theorem, $E=\pi(S)$ has Lebesgue measure zero in $W$: Sard's theorem applies in our case (even if $M$ is $C^1$-smooth) since $M$ and $W$ have the same dimension.  In other words, $E=\pi(S)$ has zero Lebesgue measure in $R^m$.   
To conclude: The interior of $N$ has full (and positive!) measure in $N$, i.e. there exists a subset $E\subset N$ of zero measure such that $N-E$ is nonempty and open in $R^m$. 
