As usual in such problems, we consider two specific cars as one block/unit.


So, now we have circular permutations of 4 objects.

The number of such circular permutations is (4-1)! = 3! = 1*2*3 = 6.


Also, there are 2 permutations inside the block, giving 2*6 = 12 different/distinguishable 
arrangements.


It is how everything works for circular permutations.


But in our case, the parking places are  .


Therefore, for 4 objects, we should multiply 12 by 4, because with numbered places,
we can start from any of 4 places from 1 to 4, which gives 4 times as many arrangements
as for the simple circular permutations case.


Thus the final answer is 4*(2*3!) = 4*12 = 48 different distinguishable arrangements.    ANSWER.


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                |   Another solution - another way of reasoning.   |
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For the two specific paired cars, we have 4 possible positions in the circular lot of 5 places.

We also have 2 possible permutations inside the pair.


For each such configurations of the two specific cars, we have (5-2)! = 3! = 6 possible permutations 
of the remaining 3 cars in 3 remaining positions.


The product 4*(2*3!) = 4*(2*6) =  4*12 = 48  gives the same answer 48.