The Dark Knight has a problem with the wedge

Let M = 1.7 kg, m = 0.6 kg
On A, parallel to the horizontal a force N cos 45° acts towards the left. Here N is the contact force between A and B which acts perpendicular to the face of the incline. So if a₀ be A's acceleration towards the left, then
N sin 45° = M a₀ ----- (1)

Taking the reference frame as the wedge A, the forces acting on B are:
mg vertically downwards,
N perpendicular to the face pushing B away,
the pseudo force ma₀ towards the right.

In this reference frame, B slides down the incline. So we have
mg sin 45° + ma₀ cos 45° = ma' -----(2)
(here a' is the acceleration of B w.r.t. A)
Normal to the face: mg cos 45° = N + ma₀ sin 45° -----(3)

Solving for a₀, a' we get
a₀ = mm+2 M g = .64 g =320 g

And a' = √2 (m+M)(m+2M) g = 23√240 g
Hence, the vertical component of B's acceleration is a' sin 45° = 2340 g

And the horizontal acceleration of B:
a' cos 45° - a₀ = 2340 g - 320 g = 1720 g

So a), b), c) are correct.

Of course, one can solve it from the ground frame entirely without reverting to any pseudo force. However, we must note that in the ground frame, the direction of B's motion won't be parallel to face of the wedge. In fact, the following vector equation holds:
\vec{a}_{BG}=\vec{a}_{BA}+\vec{a}_{AG}
(Here, a_BG denotes the acceleration of the block B w.r.t. the ground, a_BA is the acceleration of B w.r.t. A and a_AG is the acceleration of A w.r.t. the ground.)

From the vector diagram, we have
\tan 45^\circ=\dfrac{a_v}{a_0+a_h}
where a_v, a_h, a₀ are, respectively, the vertical, the horizontal accelarations of B and the acceleration of A.
Further, the net horizontal force being zero, we get Ma₀ = ma_h. These equations can be used to determine the required quantities in the ground frame.