Is this possible when it comes to the relations of P, NP, NP-Hard and NP-Complete?
Actually, your version is correct and Wikipedia's is wrong! (Except that it has a tiny disclaimer at the bottom.)

If $\mathrm{P}=\mathrm{NP}$, Wikipedia claims that every problem in $\mathrm{P}$ is $\mathrm{NP}$-complete. However, this is not true: in fact, every problem in $\mathrm{P}$ would be $\mathrm{NP}$-complete, except for the trivial languages $\emptyset$ and $\Sigma^*$.

You can't many-one reduce any nonempty language&nbsp;$L$ to $\emptyset$, because a many-one reduction must map "yes" instances of&nbsp;$L$ to "yes" instances of&nbsp;$\emptyset$, but $\emptyset$&nbsp;has no "yes" instances. Similarly, you can't reduce to&nbsp;$\Sigma^*$ because there's nothing to map the "no" instances to.  However, if $\mathrm{P}=\mathrm{NP}$, then every other language in&nbsp;$\mathrm{P}$ is $\mathrm{NP}$-complete, since you can solve the language in the reduction.

So, just to make it explicit:


your diagram is correct;
Wikipedia's isn't (unless you read the tiny disclaimer);
the area you've labelled "$\mathrm{P}$, $\mathrm{NP}$" contains the two languages $\emptyset$ and&nbsp;$\Sigma^*$, and nothing else;
the area you've labelled "$\mathrm{P}$, $\mathrm{NP}$-complete" contains every other language in&nbsp;$\mathrm{P}$ and nothing else.