db.a.drop()
db.a.createIndex({a:1, "b.$**":1, c: 1})
db.a.insertOne({})
db.a.explain().find({a: 3, 'b.x': {$ne: null}, c: 5})

This results in a plan with an index scan with an IndexBounds with two $_path keys. It seems mongosh only shows the last occurance, while the legacy mongo shell shows the first occurance. But you can set eg. a breakpoint at the plan ranker and evaluate statusWithMultiPlanSolns.getValue()[0]->toString() to see both the keys. This gives (ignore the strikethrough formatting):

(std::string) "FETCH\n-{}ns = test.a\n{}filter:\n        $and || Selected Index #1 pos 0 combine 1\n            $not\n                b.x $eq null\n            c $eq 5\n{}nodeId = 2\n{}fetched = 1\n{}sortedByDiskLoc = 0\n{}providedSorts = {baseSortPattern: {}, ignoredFields: [a]}\n{}Child:\n{}--{}IXSCAN\n{}-----{}ns = test.a\n{}-----{}indexName = a_1_b.$**_1_c_1\n{}-----{}keyPattern = { a: 1, $_path: 1, $_path: 1, c: 1 }\n{}-----{}direction = 1\n{}-----{}bounds = field #0['a']: [3, 3], field #1['$_path']: [MinKey, MinKey], [\"\", {}), field #2['$_path']: [MinKey, MaxKey], field #3['c']: [MinKey, MaxKey]\n{}-----{}nodeId = 1\n{}-----{}fetched = 0\n{}-----{}sortedByDiskLoc = 0\n{}-------providedSorts = {baseSortPattern: {}, ignoredFields: [a]}\n"

 
For some context: wildcard indices use MinKey in the $_path field to denote an object where the wildcard object is empty/absent. In this case the query specifies $ne: null, so I don't think the extra point scan on MinKey is needed. The created plan also has a residual filter in this case, so this should not lead to a correctness issue.