The problem as stated now is solvable in linear time.
To see this, suppose $p\in P$ is such that there are $x\in X$ and $w\in W$ with $p_i=x_iw_i$ for all $i$. This means on the one hand that $1=\sum_{i=1}^np_i=\sum_{i=1}^nx_iw_i$, but on the other hand because of $1 = \sum_{i=1}^nx_i$ and $1 = \sum_{i=1}^nw_i$ we have $1 = \sum_{i=1}^n\sum_{j=1}^nx_iw_j$, so that $x_iw_j$ must be $0$ for all $i\neq j$ (using also that all the $x_i,w_i$ are nonnegative).
This implies that there is some index $k$ with $x_k=w_k=1$ and $x_j=w_j=0$ for all other $j$: indeed, since $1 = \sum_{i=1}^nx_i$, some $x_k$ must be nonzero. Then $x_kw_j=0$ means $w_j=0$ for all $j\neq k$, and because $1 = \sum_{i=1}^nw_i$, we have $w_k=1$. Analogously we get $x_j=0$ for $j\neq k$ and $x_k=1$. As a consequence, $p_k=1$ and $p_j=0$ for $j\neq k$.
The decision problem is whether this is true for all $p\in P$. Since $P$ is convex, this is only possible if $P$ only consists of one of these unit vectors, i.e. for some $k$ we have $a_k=b_k=1$ and $a_j=b_j=0$ for all other $j$; additionally, $X$ must also contain this vector, i.e. we need to have $u_k=1$ and $l_j=0$ for $j\neq k$.
EDIT: What we have are the following results:
- If $p,x,w\in[0,1]^n$ with $\sum_ip_i=\sum_ix_i=\sum_iw_i=1$ and $p_i=x_iw_i$ for all $i$, then $p=x=w=e_k$ for some $k$, where $e_k$ is the unit vector given by $(e_k)_k=1$ and $(e_k)_j=0$ for all $j\neq k$.
- Adding the constraints $a_i\le p_i\le b_i$ and $l_i\le x_i\le u_i$ further restricts the set of solutions to those $e_k$ which are both in $P$ and in $X$, i.e. such that $b_k=u_k=1$ and $a_j=l_j=0$ for $j\neq k$.
- The decision problem given in the question is whether such $x,w$ exist for all $p\in P$; the answer to this is "yes" if and only if $P$ is a singleton set $\{e_k\}$ for some $k$, i.e. $a_k=b_k=1$ and $a_j=b_j=0$ for $j\neq k$, and additionally this $e_k$ is in $X$, i.e. $u_k=1$ and $l_j=0$ for $j\neq k$. This can be checked in linear time.
