I don't know anything about your schema, but if your List calculation is working then this should do the trick:
Calculation field, result text
If ( PatternCount ( CommitteList_L ; "CommitteeName" ) ≥ 1 ; "Your message would go here." ; "" )
Note the message can also be data from a field. You can also have multiple committees by using the case function.
For that calculation, I suggest a slight change:
If ( ValueCount ( FilterValues ( CommitteeList_L ;"CommitteeName" ) ) ; "Your message" ; "" )
In some cases, patterncount can return a non zero value because the second value "committeName" matches a portion of a value in the list.
A name of John Smith, for example would match to the name John Smithson and produce a non zero pattern count. This will not occur with filtervalues.
Thank you, both Bumper AND Phil. I actually chose Bumper's answer as best because my list included committee names and sub-committee names, but I wanted to test for just a committee name. I had calculated a new field prior to listing that concatenated committee_name and committee_role. Testing for just "Sunday School" wouldn't work using FilterValues since now the list read "Sunday School: Teacher" or "Sunday School: Staff," etc. Perfect instance of where PatternCount is useful. However, I can totally see why Phil's solution is good to know about, so I'm extremely grateful to be aware of that option as well. Thank you so much!