Answer

Let us reason out completely and not just “guess” at an answer.

We use base 10.

abcd = a*10^3 + b*10^2 + c*10^1 + d
and
dcba = d*10^3 + c*10^2 + b*10^1 + a

abcd x 4 = dcba implies,

4 x [ a*10^3 + b*10^2 + c*10^1 + d ]= d*10^3 + c*10^2 + b*10^1 + a

Now, 4a < 10. Why? if 4a > 10 then the LHS > 10^4 [ being a 5 digit number ] while the RHS will be <10^4 [ being a 4 digit number ] contradicting the equality above.
The only multiples of 4 that are less than 10 are 4 and 8. Hence 4a = 4,8 giving, a = 1 or 2.

Note also that 4d must ‘end’ in a ( by considering the last digits ). Also 4a must end in d ( by considering the the first digits ) in the above equation.

This means that if a =2, then d =8 [ obtained by equating the coefficient of 10^3 in the equation above ], but 4d = 32 ends in ‘a’ or ‘2’ as well and thus this could be a solution. On the other hand, if a =1 then 4a = 4 and thus d = 4 [ obtained by equating the coefficient of 10^3 in the equation above ]. But
4d = 16 and does NOT end in ‘a’ or ‘1’ as described in the earlier paragraph. Hence a =2 and d =8.

Our equation now becomes,

4 x [ 2*10^3 + b*10^2 + c*10^1 + 8 ]= 8*10^3 + c*10^2 + b*10^1 + 2

which reduces to

4 x [ b*10^2 + c*10^1 + 8 ]= c*10^2 + b*10^1 + 2

400b + 40c + 32 = 100c + 10b +2

390b + 30 = 60c

13b + 1 = 2c …(1)

Consider (1). Remember that b and c are all integers between 0 to 9 because they are digits of an integer.
2c < 2 (10) = 20.
13b + 1 < 20 using (1) and the above.
b < 19/13 = 1.46…
meaning b = 0 or b =1 since b is an integer less than 1.47.

If b = 0, equation (2) gives us that 2c = 1 or c =1/2 ( a contradiction since c is not integer ).

Hence b = 1. equation (2) gives us 14 = 2c implying that c=7.

To recap, we have found the values of a,b,c,d as follows:

(a,b,c,d ) = ( 2, 1, 7, 8 )

Indeed to check :

2178 x 4 = 8712