Regex how to match an optional character

Use

[A-Z]?

to make the letter optional. {1} is redundant. (Of course you could also write [A-Z]{0,1} which would mean the same, but that’s what the ? is there for.)

You could improve your regex to

^([0-9]{5})+\s+([A-Z]?)\s+([A-Z])([0-9]{3})([0-9]{3})([A-Z]{3})([A-Z]{3})\s+([A-Z])[0-9]{3}([0-9]{4})([0-9]{2})([0-9]{2})

And, since in most regex dialects, \d is the same as [0-9]:

^(\d{5})+\s+([A-Z]?)\s+([A-Z])(\d{3})(\d{3})([A-Z]{3})([A-Z]{3})\s+([A-Z])\d{3}(\d{4})(\d{2})(\d{2})

But: do you really need 11 separate capturing groups? And if so, why don’t you capture the fourth-to-last group of digits?