Approach using the pressure vs ln(r) profile:
A convenient way to picture near wellbore damage is to plot flowing pressure $p$ against $\ln r$ for steady radial flow. Without any damage, this plot is a single straight line all the way from $r_w$ to $r_e$, because the slope of $p$ versus $\ln r$ depends only on $q\mu/(kh)$, which stays constant when the permeability is uniform.
Introducing the damaged ring:
When the zone from $r_w$ to $r_s$ has reduced permeability $k_s < k$, the actual pressure profile has two straight segments on the $p$ versus $\ln r$ plot. The segment from $r_s$ to $r_e$ has the normal (shallow) slope set by the undamaged permeability $k$, exactly as if there were no damage at all. The segment from $r_w$ to $r_s$ has a steeper slope, since it is controlled by the lower permeability $k_s$, so the pressure falls off faster per unit of $\ln r$ inside the damaged ring.
Reading off the additional drop $\Delta p(r)$:
If we extend the undamaged (shallow slope) line backward from $r_s$ down to $r_w$, it gives the pressure that would have existed at the well with no damage. The additional drop $\Delta p(r)$ at any radius $r$ inside the damaged ring is simply the vertical gap between the actual (steeper) profile and this extended undamaged line. At $r = r_e$ down to $r = r_s$, the actual profile and the extended undamaged line are identical by construction, so the gap is zero all along that stretch, including right at $r_s$.
Growth of the gap inside the damaged ring:
Moving from $r_s$ inward to $r_w$, the actual (steeper) line drops faster than the extended undamaged line, so the vertical gap widens continuously and reaches its maximum value, equal to the classical skin drop $\Delta p(r_w) = \dfrac{141.2\,qB\mu}{kh}S$, exactly at the wellbore face.
Final Answer:
\[ \boxed{\Delta p(r_s) = 0} \]