$\angle z = \angle x + \angle y$
This is a triangle property that tells us that "the exterior angle of a triangle is the sum of the two opposite interior angle."
Similarly,
$\angle y = \angle a + \angle b$
By substituting, we get,
$\angle Z = \angle x + \angle a + \angle b$.
Now $\angle Z = \angle BCD$
$ \angle b = \angle BAD \\ \angle BCD - \angle BAD =\angle Z - \angle b \\ \qquad \qquad =\angle x + \angle a + \angle b - \angle b \\ \qquad \qquad =\angle x + \angle a \\ \qquad \qquad = \angle DEC + \angle BFC $
So, option A is the answer.