How to add optional parameters in a django view keeping the same url

Question

Let's say I have the following URL:

http://127.0.0.1:8000/en/project_page/1/

and my urls.py:

path(
    'project_page/<int:pid>/', 
    views.project_page,
    kwargs={'approvalid': None},
    name='project_page'
),

Generally speaking, the users will always access via http://127.0.0.1:8000/en/project_page/1/ , but in some cases I want to pass an optional parameter so that the page renders optional content.

For example, if the user access: http://127.0.0.1:8000/en/project_page/1/somevalue

I would like to see the content of somevalue in my view, like:

def project_page(request, pid, somevalue):
    if somevalue:
        #do something here with the optional parameter
    else:
        #run the view normally
    return render(request, 'mypage.html', context)

Once we get the optional parameter, I want the url structure to be the original:

http://127.0.0.1:8000/en/project_page/1/somevalue -> get the value of the optional parameter and get back to the original url -> http://127.0.0.1:8000/en/project_page/1/

What I've tried:

I thought kwargs would do the trick, but I did something wrong.

I found this from the official docs (Passing extra options to view functions) but it seems that I'm doing something wrong.

EDIT:

def project_page(request, pid, somevalue):
    context = {}
    context['pid'] = 'pid'
    if approvalid:
        context['nbar'] = 'example1'
        #some logic here
    else:
        context['nbar'] = 'example2'
        #some logic here
    return render(request, 'mypage.html', context)