Assuming that all matrices are n x n and invertible, solve for d .

c b (a^t) d b (c^t) a = c b^t

First multiply by C inverse on the leftC−1CBATDBCTA=C−1CB?
Yielding:BATDBCTA=BT

Then Multiply by B inverse and A transpose inverse on the left yielding
DBCTA=B−1A−TBT

Now, multiply by B inverse, C transpose inverse and A inverse on the right
D=B−1A−TBTB−1C−TA−1

Letter c i’d say