Hi,
Thank you for contacting us and I'd be happy to assist.
1. Assuming that the slug of the checkbox field is "show-child-pages" you can use the item attribute, to get the value from the parent page, like this:
( ref: https://toolset.com/documentation/programmer-reference/views/views-shortcodes/item-attribute/ )
[types field='show-child-pages' output='raw' item='$parent'][/types]
The same shortcode can be used in the conditional block's advanced editor, to see if the value is equal to '1':
( example screenshot: hidden link )
( ( ' [types field="show-child-pages" output="raw" item="$parent"][/types]' eq '1' ) )
As a result, the blocks inside this conditional block will only show if the parent page is set to show the child pages.
2. You can create a single classic/legacy editor view to cover both cases:
case a). To show the current page's child pages
case b). To show all the child pages which share the same parent page as the current child.
You'll create a classic/legacy view that shows the pages, and set the post parent query filter with the option "Post with ID set by the shortcode attribute: wpvchildof".
For case a, you'll pass on the current page's ID in the view's shortcode, like this:
[wpv-view name="view-name" wpvchildof="[wpv-post-id]"]
And for case b, you'll be able to pass on the current page's parent ID in the view's shortcode, like this:
[wpv-view name="view-name" wpvchildof="[wpv-post-id item="$parent"]"]
I hope this helps and please let me know if you need any further assistance around this.
regards,
Waqar