Overflow condition for stack suggests that stack is full.
As mentioned in the questions , the stacks are arranged from left to right in a contiguous manner i.e. after stack S_{0} , say ends ,stack S_{1} begins.
So the ith stack is full if top of ith stack points to the last element of that stack and just after that the next stack contents begin i.e. we reach base address of next stack.
So if we have ith stack the condition for it to be full is that top[i] is just before base[i+1] i.e just before base address of (i+1)th stack
Hence , the required condition is : Top[i] = Base[i+1] - 1
Hence , C) should be the correct option.
If asked for underflow , i.e. stack is empty , in that case , Top[i] = Base[i] - 1