Using the flowchart, we can solve the given question as follows:

$(5)_{10}$ = 0101(in 2’s complement)

$(7)_{10}$ =0111(in 2’s complement)

Multiplicand (B) = 0101

Multiplier (Q) = 0111

And initially Q-1= 0

Count =4

The result to the problem is $(0010 0011)_2$ or $(35)_{10}$

Here a property of Arithmetic right shift can be seen. When after addition or subtraction(A-B or A+B) a right shift occurs,the left most bit of $A(A_{n-1})$ is shifted to the second left most bit i.e. $A_{n-2}$ the $(n-1)^{th}$ bit is retained as shown in the above example. This is done to preserve the sign of A.