A wedge of mass M = 5 kg stands on a smooth horizontal floor…

Step 1. Who moves how. Let the wedge's acceleration be $A_w$, to the right, and let $b$ be A's acceleration down the slope relative to the wedge. The string has a fixed length, so B rises with $b$ relative to the wedge. B also moves sideways with the wedge, because it stays against the vertical face.

Step 2. Momentum sideways. The smooth floor gives no horizontal force, and the system starts at rest, so its horizontal momentum stays zero. Over the ground, A's horizontal acceleration is $A_w - b\cos\theta$, while the wedge and B both have $A_w$:

Step 3. Block A, in the wedge's frame. A feels a pseudo force $m_A A_w$ to the left. Its part down the slope is $m_A A_w\cos\theta$. Along the slope:

Step 4. Block B. Vertically: $T - m_B g = m_B b$.

Step 5. Solve. Add the equations of Steps 3 and 4, and put in $A_w$ from Step 2:

Then $A_w = \dfrac{4\ \text{kg} \times 3.52\ \text{m/s}^2 \times 0.8}{10\ \text{kg}} = 1.13\ \text{m/s}^2$.

Step 6. The forces. $T = m_B(g + b) = 1\ \text{kg} \times (10 + 3.52)\ \text{m/s}^2 = 13.5\ \text{N}$. B is pushed sideways only by the vertical face, so $N_B = m_B A_w = 1.13\ \text{N}$. It is positive, so B really stays pressed against the face.