Implication of Berman and Hartmanis conjecture
Two languages $X$ and&nbsp;$Y$ are $p$-isomorphic if there's a polynomial-time computable bijection $f\colon X\to Y$ such that $w\in X\iff f(w)\in Y$.

Now, suppose that P$\,=\,$NP.  Assuming we're defining NP-completeness in terms of polynomial-time many-one reductions, every language in NP, except for $\emptyset$ and $\Sigma^*$ is NP-complete.  To see this, let $Y\in\mathbf{NP}\setminus\{\emptyset, \Sigma^*\}$ and pick two strings $w_{\mathrm{in}}\in Y$ and $w_{\mathrm{out}}\notin Y$. For any language $X\in \mathbf{NP}$, we can reduce $X$ to&nbsp;$Y$ in polynomial time using the function
$$f(w) = \begin{cases}
\,w_{\mathrm{in}} &amp;\text{if }w\in X\\
\,w_{\mathrm{out}} &amp;\text{if }w\notin X\,.
\end{cases}$$
Since $X$ can be decided in polynomial time, we can compute $f$ in polynomial time.  Note that there is a small error in the proof quoted in the question, which claims that all languages in P would be NP-complete if P$\,=\,$NP.  This is incorrect: if we took $Y=\emptyset$ or $Y=\Sigma^*$, the reduction wouldn't work because one of $w_{\mathrm{in}}$ and&nbsp;$w_{\mathrm{out}}$ wouldn't exist.

So, we've established that, if P$\,=\,$NP, then every finite language is NP-complete; we also know that some infinite languages, such as $\mathrm{SAT}$ are NP-complete.  But now suppose that there's a $p$-isomorphism between every pair of NP-complete languages.  That means that, in particular, there's a bijection between every finite language and $\mathrm{SAT}$. But there can't be a bijection between a finite set and an infinite set.