If y = a(x - b)² + p intersects x-axis, then the intersection point will be of the form (k, 0)

So, 0 = a(k - b)² + p
So, (k - b)² = -p/a

Now, (k - b)² cannot be negative.
So, if -p/a is non-negative, then only the graph of the given equation can have intersection with x-axis.

So, the problem is basically asking "Is -p/a ≥ 0?"

Clearly, none of the statements are individually sufficient but both of them together is sufficient.

Answer : C