Create a method that accepts 4 PVector objects to represent the corners then inside this method call the drawLine() method created by James four times one for each side.

Something like should do the trick.

Code:

void drawBox(PVector c1, PVector c2, PVector c3, PVector c4, float weight, int strokeColour){
   drawLine(c1, c2, weight, strokeColour);
   drawLine(c2, c3, weight, strokeColour);
   drawLine(c3, c4, weight, strokeColour);
   drawLine(c4, c1, weight, strokeColour);
}

// method created by James Carruthers (parameters have been 
// changed to PVectors and 'color' to int)
void drawLine(PVector p1, PVector p2, float weight, int strokeColour)
{
  PVector v1 = PVector.sub(p2, p1);
  float rho = sqrt(pow(v1.x,2)+pow(v1.y,2)+pow(v1.z,2));
  float phi = acos(v1.z/rho);
  float the = atan2(v1.y,v1.x);
  v1.mult(0.5);
  
  pushMatrix();
  translate(x1,y1,z1);
  translate(v1.x, v1.y, v1.z);
  rotateZ(the);
  rotateY(phi);
  noStroke();
  fill(strokeColour);
  box(weight,weight,p1.dist(p2)*1.2);
  popMatrix();
} 

Alternatively since a rectangle with thickness is a box you can use the Shapes 3D library and the Box class now you're used to Tubes.

When you say thickness, what exactly do you mean?  Do you mean the box should rise above the four corner-points, or below, or be centered around those points?  

Also, to define a plane, you only need three points.  Since you want draw a rectangular portion of the plane, three points are also enough to figure out all of the edges of the rectangle.  In general the shape will be a parallelogram instead of a rectangle, unless you are careful to make the edges perpendicular when determining the three points.

Four points aren't necessarily all on the same plane because the fourth point might not be on the plane defined by the other three points.  Four points basically make two triangles that are joined on one edge, that may or may not be on the same plane.  Are you making sure that the four points are actually all on the same plane?

The approach I would take for this is to take the three points and find the two edges of the rectangle that are between these three points.  These will be two vectors.  Now find the cross product between those two vectors.  This will give you a vector perpendicular to the plane.  Normalize it, and then multiply it by half of the thickness.  Let's call that vector v. You need the fourth point now (which I'll assume you already have).  Now take each of the four points, and define the four points that make up the top face by adding v to your original four points, and then define the four points that make up the bottom face by subtracting v from your original four points.  Now you have all of the points you need to draw the six faces of the box.

Basically you seem to want the equivalent of a piece of paper. You might look at the 3rd from last example beginShape(QUADS)) on this page

Ok so its 2D but vertax can also be take 3 values x/y/z when using P3D or OPENGL

Not forgetting what d.rifkin said about 4 points not being on a plane.