Extensional flow of a long thin thread

The term you are referring is from using absolute pressure rather than gauge pressure, and is the result of integrating the external pressure vectorially over the external surface from z to z + dz. If the area were not changing, the pressure contribution would have no axial component, and would be zero.

The 2nd equation is just from the 3D version of Newton's law of viscosity. The p in this equation is not the external pressure, and is determinate only from considering the boundary conditions (and is equal to $p_{ext}$ only at the boundary. For an incompressible fluid, it is not the same as the pressure determined from the equation of state, and only represents an isotropic contribution to the stress tensor.

INTEGRATION OF EXTERNAL PRESSURE OVER SURFACE

The external pressure acts normal to the filament surface. The unit outwardly directed normal to the filament surface is given by $$mathbf{n}=frac{mathbf{i_r}-frac{dr}{dz}mathbf{i_z}}{sqrt{1+left(frac{dr}{dz}right)^2}}$$So, the force per unit area that the external pressure exerts on the filament surface is $$-P_{ext}mathbf{n}=frac{-mathbf{i_r}+frac{dr}{dz}mathbf{i_z}}{sqrt{1+left(frac{dr}{dz}right)^2}}P_{ext}$$The radial component of this force per unit area integrates to zero, and the axial component in the positive z direction is $$frac{frac{dr}{dz}}{sqrt{1+left(frac{dr}{dz}right)^2}}P_{ext}$$The differential area of surface upon which this force acts between z and z + dz is $$2pi rsqrt{1+left(frac{dr}{dz}right)^2}dz$$So the axial component of force that the external pressure exerts on the filament between z and z + dz is $$2pi rfrac{dr}{dz}dzP_{ext}$$So, dividing by dz in the force balance gives $$P_{ext}2pi rfrac{dr}{dz}=P_{ext}frac{dA}{dz}$$

As far as the derivation of the 2nd highlighted equation is concerned, the development is too extensive to present here. For the full. derivation of this tensorial equation, see Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 1