您可以尝试：

(?:^letters=|\G(?!^))[BC]*\K[AD-Z]+

查看在线demo

• (?:^letters=|\G(?!^))-非捕获组，用于匹配输入的开头，如字面上的"letters ="* 或 * Assert上一匹配结尾处的位置，但否定字符串的开头;
• [BC]*-0+乘以字母"B"或"C";
• \K-重置上报匹配的起始点;
• [AD-Z]+-匹配范围"AD-Z"中的1+个大写字母。

替换为空字符串：

echo preg_replace('/(?:^letters=|\G(?!^))[BC]*\K[AD-Z]+/', '', $string);

如果我正确理解了这个问题，你可以用一个空字符串替换下面正则表达式的每个匹配项。

^.*'(?=(?:letters|LETTERS)=[A-Z]+';$)|[AD-Z](?!.*=)|';$

Demo
这个表达式可以分解如下。

^             match beginning of string
.*            match zero or more chars other than line terminators
'             match literal
(?=           begin a positive lookahead
  (?:         begin a non-capture group
    letters   match literal
  |           or
    LETTERS   match literal
  )           end the non-capture group
  =           match literal
  [A-Z]+      match one or more uppercase letters
  ';          match literal
  $           match end of string
)             end the positive lookahead 
|             or
[AD-Z]        match a capital letter other than 'B' or 'C'
(?!           begin a negative lookahead
  .*          match zero or more chars other than line terminators
  =           match literal
)             end the negative lookahead
|             or
';            match literal
$             match the end of the string