Suppose $M$ is a smooth manifold and $A$ is a $\mathcal{C}^\infty(M)$-submodule of $\Gamma(TM)$ with the property that for all $x \in M$, the values of the vector fields in $A$ evaluated at $x$ determine a $p$-dimensional subspace of $T_x M$. Then $A$ defines a smooth distribution $P$ i.e. a smooth subbundle of $TM$ of rank $p$; for every $x \in M$, $P_x$ is given by all the vector fields in $A$ evaluated at $x$.
Consider now $T_P \subset \Gamma(TM)$ the $\mathcal{C}^\infty(M)$-submodule of vector fields which are tangent to $P$ i.e. their values are in $P_x$ for each $x \in M$. In other words $T_P = \Gamma(P)$.
My question is, why is it that $A = T_P$? A priori, it only seems that $A \subset T_P$, but why is it that if a vector field is in $P_x$ for every $x$ then it must be in $A$?
As a preliminary step, it seems to me that it is not necessary for $A$ to be closed under multiplication with smooth functions for it to define a distribution. Indeed, if $A \subset \Gamma(TM)$ is just a collection of vector fields which at every point span a $p$-dimensional subspace, then for every point we find some vector fields in $A$ which are linearly independent at that point, so by continuity of $\det$ they must be linearly independent on a neighborhood of the point, so these vector fields smoothly trivialize $P$ around the point, hence $P$ is a smooth distribution. Am I wrong with anything in this argument?
Based on it, it seems that the fact that $A$ is a $\mathcal{C}^\infty(M)$-submodule is essential in proving $T_P \subset A$, but I can't figure out how to use this fact.