\somepenalty=10000 vs. \somepenalty 10000 vs. \somepenalty10000

There is no difference. The syntax for a primitive tex assignments allows an option = (and optional space)

Note that

 \interlinepenalty 10000 or \interlinepenalty10000

are the same for a different reason, the space after a command name is never tokenized at all it just terminates the command so these make the same tokens.

Conversely

\interlinepenalty= 10000 

the space after = is tokenized but is ignored as the assignment absorbs optional equals optionally followed by a space.

David C. explained that there is no differences between these three alternatives because of TeX syntax rules (the optional =) or because of the toknizer outputs the same result in case 2 and 3. But you have asked why? The reason is: The optional = is used because this is more understandable by humans. In old days (when TeX was born) each byte was counted. So: the = was not used in the macro bodies because these token strings must be saved into the memory. But this optional = was used outside macro bodies. Today, there is no such memory limitations, the usage of = is recommended.

Your second question: is there any case when = cannot be used. Yes. If the syntax context is not assignment. For example \penalty 10000. The primitive command \penalty is not primitive register, so this is not assignment but this is command to do something in vertical or horizontal list. You cannot use = here.

My last comment: the optional = may have optional space around it (one space left and one right of the =. So, \interlinepenalty= 10000 is also possible. And

\interlinepanalty       =     10000

is also possible, because there is no space before = and only one space after the = character because the tokenizer converts this to \interlinepenalty= 10000.

Unless I overlooked it within the answers given already, one open question remained: are there cases where the = is required? Yes, for example if you want to do

\let\foo $      % fine
\let\bar =      % incomplete
\let\baz !

In the above \bar will be let to \let and not to an equal sign, so you would need

\let\bar = =

(with or without the spaces). Even more tricky: \let something to a "space" token (exercise for the reader).