For a reversible process, it is assumed that the external pressure $P_{ext}$ is infinitesimally different from internal pressure $P_{int}$.
So in reversible process, I can have $~P_{int}=P(V,T)~$ but also $~V=V(P_{int},T)~$.
Then what is the obvious reason/advantage for defining $~\delta W=-P(V,T)dV~$ instead of just define $~\delta W=-V(P,T)dP~?~$
it seems that I can always convert one to the other
Every system is defined by a variety of extensive variables: entropy, volume, mass, charge, surface area, magnetization, etc. The internal energy $U$ is a function of these variables—$U=U(S,V,N, Q,A,M,\dots)$—and we seek to know how $U$ changes when the variables change:
$$dU=\left(\frac{\partial U}{\partial S}\right)_{V,N,Q,A,M\dots}dS+ \left(\frac{\partial U}{\partial V}\right)_{S,N,Q,A,M\dots}dV+ \left(\frac{\partial U}{\partial N}\right)_{S,V,Q,A,M\dots}dN+\cdots.$$
We happen to call one of these coefficients, $\left(\frac{\partial U}{\partial V}\right)_{S,N,Q,A,M\dots}$, the negative pressure $-P$. We can increase the energy of a system for positive values of $-P\,dV$. That’s why the parameters are ordered the way they are.
in reversible process, why define $\delta W=-PdV$ instead of $\delta W=-VdP$?
You need to differentiate between work done in a closed system and an open system.
For a closed system, where there is no flow of mass into or out of the system, work is referred to as "boundary work". This is the work only associated with expanding or contracting the boundaries of the system, i.e., changing the volume of the system. The often used example is the expansion or compression of a gas in a cylinder fitted with a movable piston. For a closed system, where expansion work is considered positive,
$$\delta W=PdV$$
It applies to an irreversible as well as a reversible process. For the irreversible process, $P$ is the external pressure.
For an open system, where work is required to push or pull mass into and out of the system because of pressure differences at the boundary, work is then called flow work and
$$\delta W=VdP$$
Hope this helps.
Maybe I misunderstand your question, but pdV-work follows naturally from the definition of work $W$ from classical mechanics (in one dimension): $$ dW = F dx = pA dx = p dV $$ where $p$ is pressure, $A$ is the cross sectional area, $F$ is the applied force, $dx$ is the displacement, and $dV(=Adx)$ is the displaced volume.
Furthermore, the thermodynamic identity takes the form $$ dU = TdS - pdV $$ for internal energy. If you formulate the thermodynamic identity in terms of enthalpy you obtain $$ dH = TdS + V dp. $$ So the two terms ($pdV$ and $Vdp$) are related through the Legendre transformation, and their use depends on which potential is relevant to the problem at hand.
