Can B make a path where he's a piece of the path itself? Is it possible? I tried to make one. Why doesn't this version work? Why B doesn't make that path?
What is also interesting is how he responds if you try to tap his head. It's like his head doesn't exist and the tap appears beyond his head. Try tapping with different screen perspectives (i.e. rotate the screen) to see. Therefore, my opinion is that B (and whatever head he has) are expressly programmed to not factor into any route planning. Starting with the tap itself, it's like B isn't even there - he merely responds to commands given him.
As said before, There isn't a way to make a path where B is directly a piece of the path itself. But there is a way to make it where B is a support (undirectly) . It is possible using a motor's particularity (that has been discovered by @Gepeto, and @Block builder did a knockoff of his demo).
And now, it's a level. http://mekoramaforum.com/media/balloon-boy.8881/ Such ideas can make for better levels with a sense of puzzle. But having it as a single entity is just too unstable. So that will have to be separate stuff on B.
An unique piece is a group of parts. Each block attached directly to B,obtains his proprieties. "Motor block" has a particularity: to attach blocks,but they won't obtain proprieties each other. So that piece (with B,R,1 zap,2 motors) is unique but the top doesn't obtain B's proprieties(especially that he can't make a path where he's a piece of the path itself). So it's an unique piece. That's the only way to make a path where [...] itself. Can B make a path where he's a piece of the path itself? Yes.
I think some will still argue that a zapper or zapperbot is the path finishing thing. Whether it is a part or a piece. But I consider R+B a single thing when attached. But another question, does that thing still define as "B"?
The title of this thread is "Can B make a path where he's a piece of the path itself?" @fire_rabbit perhaps that is what you meant by "that's not as expected," but in your example, B is not a part of the path. True, B is supporting P, but P could equally be on a slider or some balls that roll away; B himself does not constitute a step in the path calculation to the win.
This version by @Logo doesn't answer to the requisite "part directly attached with B" but it's the most simple one I saw: Spoiler: Card
