Can evaluation be enough here ?

[xr_a_y]

How can this be solved ?

[d] B6k/1B6/2B4P/3B2KP/4B2P/5B2/6B1/7B w - - 0 1

[mar]

maybe, you'll never get this position OTB, though
what might work is to count all bishops on same color as one (material-wise)
then you have 3 wrong color pawns wrt to the bishop, so they won't promote

something like this. of course in cases where the pawn would escape and queen, the search will take care of that

[RubiChess]

8 bishops with 3 pawns still on board? I would evaluate it as "illegal position".

[xr_a_y]

This one is more "legal"

[d] 7k/8/7P/6KP/7P/8/6B1/7B w - - 0 1

I guess, indeed the idea to "fuse" same color bishop seems promising...
Easy for bare king case, harder for K+Ps cases ...

For instance, of course

[d] 7k/8/7P/p5KP/p6P/p7/6B1/7B w - - 0 1

and

[d] 7k/8/7P/6KP/7P/ppp5/6B1/7B w - - 0 1

won't have the same outcome

[peter]

How can this be solved ?

[d] B6k/1B6/2B4P/3B2KP/4B2P/5B2/6B1/7B w - - 0 1

There are two white bishops too much on the board for a legal position.