Proof of $\det(\textbf{ST})=\det(\textbf{S})\det(\textbf{T})$ in Penrose graphical notation
Solution 1:
(see explanatory note if confused)
Penrose writes "The antisymmetrizing bar can be inserted in the middle term because there is already antisymmetry in the index lines that it crosses." in the caption for Figure 13.8
What he means is that "antisymmetrization" is an idempotent tensor operator. This, in turn, follows from the fact that the antisymmetrizer can be viewed as a projection onto the totally antisymmetric subspace, and the fact that projection operators are idempotent. (See Antisymmetrizer)
Recall that an operator $I$ is idempotent iff $I^2=I$. In other words, we can replace the single bar in the diagram with two bars. That is all.
Explanatory Note: I have not posted images because (a) they are under copyright and (b) they are not relevant to the crux of the question.
I looked at the google cache of this webpage and noticed that all the attached diagrams were screenshots from the book "The Road to Reality" by Roger Penrose, which I have access to.
Figures 12.17 and 12.18 are located on pages 241 and 242, respectively.
The proof KalEl is referring to is found as Figure 13.8, on page 261.