Let $P$ be the number of distinct ways of painting six sides of a cube with 6 distinct colours, where you would regard two cube colourings the same if one can be obtained from the other by rotation, then sum of digits of  $P$  is ________

Colour one side with the ugliest colour, and put the cube on a table ugly side down. There are 5 choices for the colour on top. For each of these choices, colour the side facing you with the nicest remaining colour. The last three sides can be coloured in $3!$ ways, so the number of colourings is $\left(5\right)(3!).$