To build 16k x 16 RAM, we need 16x2=32 number of 1k x 8 RAMs, arranged in 2 columns of 16. To select one of those chips from a row, we need $4$ 2x4 decoders. To select one of those 4 decoders, we need $1$ more 2x4 decoder. We can use the enable input to select one of the columns of 1kx8 RAMs from 2. Thus total 2x4 decoders needed is $5$.
So the answer is (B).
The 8085 flag register will look like this:
[SUB A => I don't know what this does, but let's move on., subtract A from A, which may result in zero altogether I guess ]
MVI B,(01)H => Stores the value 01 to B
DCR B => will decrement B to zero. Let's check the flag bits now:
$D7$ : sign flag : The result is 0, so sign flag is $0$.
$D6$: Zero flag : The result is 0, so zero flag is $1$.
$D5$: Not used.
$D4$: Auxiliary Carry flag : There is no carry from lower nibble to higher nibble, so this flag is set to $0$.
$D3$: Not used.
$D2$: Parity flag : since D0 is 0, for even parity, this flag should be $1$.
$D1$: Not used
$D0$: Carry flag : No carry, so this bit is set to $0$.
Contents of the flag register will now be 0100 0100 which is $(44)H$
So, The answer is None of these.