BoundaryStyle -> None appears to be the fastest.

Yes, I found that out after I made the post. It also removes the outer boundary, so I also have to do another plot which just does the outer boundary.

I believe the problem here is that when the angle is held at 0, ParametricPlot draws the offending line segment as part of the boundary.

First, a graph just showing the segment with the angle held at 0.

p = 0;
noAngleDep = 
 ParametricPlot[{0.2178710789897841 - 
    0.9719522806619472 ss Cos[p]^3 + 0.23517815399402978 ss Sin[p]^3, 
   0.025022182511575484 - 0.23517815399402978 ss Cos[p]^3 - 
    0.9719522806619472 ss Sin[p]^3}, {ss, 0, 0.281546015800896}, 
  PlotStyle -> Green ]

Then your code:

angleDep = 
 ParametricPlot[{0.2178710789897841` - 
    0.9719522806619472` ss Cos[p]^3 + 
    0.23517815399402978` ss Sin[p]^3, 
   0.025022182511575484` - 0.23517815399402978` ss Cos[p]^3 - 
    0.9719522806619472` ss Sin[p]^3}, {p, 0, 2 \[Pi]}, {ss, 0, 
   0.281546015800896`}, PlotStyle -> Red]

And one on top of the other:

Show[{angleDep, noAngleDep}]

I believe you can set the color and opacity of the region this way:

angleDep = 
 ParametricPlot[{0.2178710789897841` - 
    0.9719522806619472` ss Cos[p]^3 + 
    0.23517815399402978` ss Sin[p]^3, 
   0.025022182511575484` - 0.23517815399402978` ss Cos[p]^3 - 
    0.9719522806619472` ss Sin[p]^3}, {p, 0, 2 \[Pi]}, {ss, 0, 
   0.281546015800896`}, PlotStyle -> {Red, Opacity[1]}]

and set the color of the "no angle dependence" plot to the same color:

p = 0;
noAngleDep = 
 ParametricPlot[{0.2178710789897841 - 
    0.9719522806619472 ss Cos[p]^3 + 0.23517815399402978 ss Sin[p]^3, 
   0.025022182511575484 - 0.23517815399402978 ss Cos[p]^3 - 
    0.9719522806619472 ss Sin[p]^3}, {ss, 0, 0.281546015800896}, 
  PlotStyle -> Red]

And then just show them together, and you should be set:

Show[{angleDep, noAngleDep}]