Suppose E, F, and G are expressions that don't involve operators of higher precedence than the comma. Are the expressions ((E, F), G) and (E, (F, G)) equivalent in scalar, list, and/or void contexts? More precisely, can you always replace ((E, F), G) with (E, (F, G)) and vice-versa without affecting program execution?
Context does not affect how code is parsed.
The comma/list operator is guaranteed to evaluate each operand from left to right (regardless of context).
From that and a couple more pieces of information, we can prove that E, F, G, ( E, F ), G and E, ( F, G ) are equivalent.[1]
Since you specifically mentioned context, we'll look at that in more detail.
For void, list or indeterminate[2] context c, we get the same context c for each item in all cases.
c c c
------- ----------- -----------
c c c c c c c
- - - -------- - - --------
c c c c
- - - -
E, F, G ( E, F ), G E, ( F, G )
For scalar context (s), we get void context (v) for all but the last item in all cases.
s s s
------- ----------- -----------
v v s v s v s
- - - -------- - - --------
v v v s
- - - -
E, F, G ( E, F ), G E, ( F, G )
I'd love to be able to say they compile identically, but they don't. Despite the documentation saying it's a binary operator, it's implemented as a n-ary operator. The parens causes another instance of the operator to be created (effectively making it a binary operator in your examples).
When it's the last expression of a sub, the context is only known at run-time, which I called "indeterminate". This makes no difference except for the the list/comma operator in scalar context. The run-time context is propagated to every operand, so you get s,s,s if it's only know at run-time, despite getting v,v,s if the context is known at compile-time.
As you can see above, this doesn't affect the answer. E, F, G, ( E, F ), G and E, ( F, G ) are equivalent whether the context is known at compile-time or not.