Google Sheets – How to Join Text if It Starts With…

I don't think rowHeight can help you here, because it refers to the entire row, not to any particular cell in it. Sorting each column by RowHeight would just rearrange the rows without making your spreadsheet more compact.

Here is the script that sorts each column by the length of cell content, ascending. The first row (headers) is left in place; empty cells are ignored.

function myFunction() {
   var ss = SpreadsheetApp.getActiveSpreadsheet();
   var sheet = ss.getActiveSheet();
   var range = sheet.getDataRange();
   var width = range.getWidth();
   var height = range.getHeight();
   var column, i, j, content;
   for (j = 1; j <= width; j++) {
     column = [];
     for (i = 2; i <= height; i++) {
       content = range.getCell(i,j).getValue();
       if (content) {
         column.push(content);
       }
     }
     column.sort(function (a,b) {return a.length - b.length;});
     for (i = 0; i < column.length; i++) {
       range.getCell(i+2,j).setValue(column[i]);
     }
     for (i = column.length; i < height-1; i++) {
       range.getCell(i+2,j).setValue("");
     }
  }
}

But in practice, this particular sort does not help all that much, because (as you noticed) the longer texts can still match up against shorter texts. Here is another sort, somewhat along the lines of what you mentioned: after sorting, each column is dropped down so that all the longest entries are in the same row. This will typically result in some empty cells at the top.

function myFunction2() {
   var ss = SpreadsheetApp.getActiveSpreadsheet();
   var sheet = ss.getActiveSheet();
   var range = sheet.getDataRange();
   var width = range.getWidth();
   var height = range.getHeight();
   var column, i, j, content;
   for (j = 1; j <= width; j++) {
     column = [];
     for (i = 2; i <= height; i++) {
       content = range.getCell(i,j).getValue();
       if (content) {
         column.push(content);
       }
     }
     column.sort(function (a,b) {return a.length - b.length;});
     for (i = 0; i < column.length; i++) {
       range.getCell(height-column.length+i+1,j).setValue(column[i]);
     }
     for (i = 2; i <= height-column.length; i++) {
       range.getCell(i,j).setValue("");
     }
  }
}

Here is an approach similar to yours; the filtering is done by regexmatch in the middle.

=countif(mmult(
   arrayformula(if(regexmatch(B3:D,"^[^MP]$"),1,0)),
   transpose(arrayformula(column(B3:D)))
),">0")

Explanation:

1. regexmatch(B3:D,"^[^MP]$") matches the cells that contain one letter which is neither M nor P.
2. if(...,1,0) converts boolean match result to 1 or 0.
3. arrayformula applies the above to the whole array, obtaining a matrix of 1s and 0s. It remains to count the number of rows that contain at least one 1.
4. As in your formula, this is done by multiplying the matrix by a vector with positive entries and counting the number of positive results. The vector I use is transpose(arrayformula(column(B3:D)))
5. After the multiplication, countif(..., ">0") counts the positive entries.