Runic Enchantments, 210 bytes

>yy `AJK`06B$̤$@
>`BCDEGPTVZ`06B$$$$ $̤$y $ $y @
>̤`IY`06Byyy$yyy̤ @
> ̤`QU`06Byy̤ $y @
> ̤`FSX`06B $yy̤$yy@
>y̤ `MN`06Byyy $@
f}l3-[r
3-[2'RA?rR1Kl'RAs]1-:0)?l
> ̤`HLORW`06Bo$y $y$y$yy@ ~B͍

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The randomization is not uniform as there isn't a good way to do that in Runic. Instead it randomly rotates each collection of letters (e.g. [BCDEGPTVZ] is one grouping) by some amount (e.g. rotating the above set by 4, where the top of the stack is at the right, the result would be [BCDEGZPTV]) and then randomly decides whether or not to reverse the stack. It performs these operations 15 times. As a result, all possible orderings are possible but not equally likely. (In the event that this is not enough, increasing it further costs zero bytes, up to 15000 shuffle loops).

This is the section of the code that handles the shuffling:

 v vvvv Loop counter

f}l3-[r < Loop entry
[2'RA?r1KRl'RAs]1-:0)?l3-
 ~{{B͍ < loop exit

 ^^^^^^ Randomly reverse
 ^^^^^ Rotate by a random an amount

The rest of the code unrolls into this:

 ABCDEFGHIJKLMNOPQRSTUVWXYZ
> `AJK`08B $ $$;
> `BCDEGPTVZ`08B $$$$ $ $ $ $ $;
> `IY`08B $ $;
> `QU`08B $ $;
> `FSX`08B $ $ $;
> `MN`08B $$;
> `HLORW`08Bo $ $ $ $ $;

^ Create IPs
 ^^^^^^^^^^^^^^^^ Set letter groupings
 ^^^ Call shuffle function
 ^^^^^^^^^^^^^^^^^^^^^^^^^^ Print letters
 ^ Last letter group needs to be sorted

If the letters are left unshuffled (but once-reversed) by changing two bytes the alphabet is printed normally, which can be used to verify that all letter groupings print in the correct places. The whitespace shifting the B commands out of phase is so that all the IPs can use the function loop at the same time without colliding, and then getting them back into phase again.

To golf, first any space that could be removed on all lines was trimmed, then each two spaces were converted to a y, and every sequence of yyyy was converted to ̤ because ̤ and yyyy are the same amount of delay, but 2 bytes cheaper. The loop exit was also combined with the HLORW main program segment in order to save on the spacing bytes (12 bytes).

Perl 5, 103 91 85 bytes

mapmy%l;@l/./g++;@k/./g=keys%lAJK,BCDEGPTVZ,SXF,IY,QU,MN;say map$k$_||$_,A..Z

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This code (ab)uses the fact that Perl's output of hash keys (%l) is random to create a mapping (%k) of all the modifiable letters to one of their possible counterparts. At output time, any key that doesn't exist is assumed to be unchanged.

Jelly, 34 bytes

ØA“wẸXWỵḲ⁻ȦƙṄṇ’œ?µ“ĠQ’ḃ3ÄœṖ⁸Ẋ€Ẏị@Ụ

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Python 3, 149 bytes

a,b,i,q,s,m,h,l,o,r,w=(set(s)for s in["AJK","BCDEGPTVZ","IY","QU","SXF","MN",*"HLORW"])
print([eval(c).pop()for c in[*"abbbbsbhiaalmmobqrsbqbwsib"]])

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Randomization by using pop() for letter set

APL (Dyalog Extended), 55 bytes

Full program. Prints uppercase with a leading and trailing space, but no intermediary spaces.

(?⍨∘≢⊇⊢)@(∊∘⍺)⊢⍵/⌈'AjkBcdegptvzIyQuSxfMn'(⊂⍤⊢,⍨∊⊂⊣)⎕A

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⎕A the uppercase alphabet

'AjkBcdegptvzIyQuSxfMn'(…) apply the following anonymous tacit function with that as right argument and the indicated string as left argument:

⊣ for the left argument,

⊂ partition it, beginning a new segment where

∊ the left arguments characters are members of the right argument (i.e. on uppercase letters)

,⍨ append

⊂ enclose (to treat it as a single element)
⍤ the
⊢ right argument

⌈ uppercase everything

…/ reduce by the following anonymous lambda, giving …"QU"λ("SXF"λ("MN"λ"A-Z")):

 ⊢⍵ on the right argument (the scrambling-in-progress alphabet)

 (…)@(∊∘⍺) apply the following anonymous tacit function to the subset that is a member of the left argument (a rhyme group)

  ⊢ on that subset

  ⊇ reorder it to be

  ?⍨ a random permutation

  ∘ of length

  ≢ tally of letters in the subset