Use two maps - one holds char freq. of pattern, the other holds char freq. of current window.

static bool findPermutation(const string &str, const string &pattern)
    {
        unordered_map<char, int> pattern_map;
        unordered_map<char, int> string_map;
        int K = pattern.size(), start = 0;

        for(char c : pattern)
          pattern_map[c]++;

        for(int end = 0; end < str.size(); end++){
          string_map[str.at(end)]++;

          if(end > K - 1){
            string_map[str.at(start)]--;
            
            if(string_map[str.at(start)] <= 0)
              string_map.erase(str.at(start));

            start++;
          }

          if(string_map == pattern_map)
            return true;
        }
        
        return false;
    }

If at any point both maps are the same, we have found the permutation.

The added map adds some space complexity, but I find this solution much easier to follow.