It doesn't, really.
You are definitely free to introduce a new vector field $p^\mu$ and add an interaction $\bar \psi \!\!\not\! p\,\psi$ to your Lagrangian, together with a kinetic term for $p^\mu$. Whether you do this or not, you must include the standard kinetic term $\bar \psi \!\!\not\! \partial\,\psi$ anyway. In this case the new term $\bar \psi \!\!\not\! p\,\psi$ does not replace the old term $\bar \psi \!\!\not\! \partial\,\psi$ but rather they appear together.
The reason the term $\bar \psi \!\!\not\! \partial\,\psi$ must always be present is the following. This term is a kinetic term, without it the equations of motion of $\psi$ do not include derivatives. In absence of kinetic terms, the field $\psi$ is essentially frozen in spacetime, with no dynamics of its own. Its Euler-Lagrange equations are algebraic equations, so that $\psi$ rigidly follows the dynamics of the rest of fields.
In the standard terminology, a field $\psi$ with no kinetic term is auxiliary, it basically acts as a Lagrange multiplier. The path integral for auxiliary fields can always be done explicitly: you just solve the classical equations of motion, and plug the result back into the Lagrangian. One says that $\psi$ is integrated out.
So, all in all, the replacement $\partial\to p$ renders the field $\psi$ non-dynamical, so it becomes an entirely different theory, one much more trivial than the original one (instead of being more general).
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Not sure where this idea is coming from, but perhaps the following will help illustrate why $\partial \to p$ is not a natural or useful operation. In quantum mechanics, operators and differential operators are emphatically not on the same footing. A regular operator is a map $\mathcal H\to\mathcal H$, a differential operator is a map $\mathrm{End}(\mathcal H)\to\mathrm{End}(\mathcal H)$. These are entirely different objects, so the proposed replacement is mostly meaningless from a mathematical point of view.
The apparent formal similarity between $\bar \psi \!\!\not\! p\,\psi$ and $\bar \psi \!\!\not\! A\,\psi$ is due to bad notation: the derivative should be denoted by $\partial$, not $p$. The differential operator maps operators into operators, a regular operator maps states to states. Again, these are completely different operations, and the notation really should reflect that. Put it differently, in $\not p\psi$, $p$ is acting on $\psi$; in $\not A\psi$, $A$ is multiplying $\psi$ (i.e., it denotes a composition). A more explicit notation would be $p(\psi)$ vs $A\circ\psi$, in which case the formal similarity breaks down.
In regular QM, the position $x$ is the eigenvalue of $\hat X$, and $\partial$ is just the matrix element of $p$ in the basis of eigenvectors of $\hat X$. In QFT, there is no $\hat X$, and $\partial$ does not denote the matrix element of some abstract operator. Instead, $x$ denotes a label on operators, with no deeper origin. The $x$-dependence of $\psi$ is fundamental, not a consequence of something else, unlike in regular QM, where $\psi(x)=\langle x|\psi\rangle$.
The analogue of $\partial$ in regular QM is $\frac{\mathrm d}{\mathrm dt}$, which is a differential operator that acts on operators. There is no abstract operator for which $\frac{\mathrm d}{\mathrm dt}$ is a matrix element. In field theory, $\frac{\mathrm d}{\mathrm dt}$ is replaced by $\partial$ and, again, there is no operator for which it is a matrix element. See also this PSE post.
