require() and include() will open the file corresponding to the path/name they receive.
Which means that, with your code, you would have to have a file called diggstyle_code.php?page=1 on your disk. That’s obviously not the case, so it fails.
Quoting the Variable scope page of the PHP Manual:
The scope of a variable is the context within which it is defined. For the most part all PHP variables only have a single scope. This single scope spans included and required files as well.
In your case, you don’t need to pass the variable. If you have a variable in your current script, it will also exist in the script you include, outside of functions, which have their own scope.
In your main script, you should have:
$page_no = 10;
require 'diggstyle_code.php';
And in diggstyle_code.php:
echo $page_no;
// Or work with $page_no the way you have to
Remember that including/requiring a file is exactly the same as copy-pasting its content at the line it’s required.