The percentage increase or decrease is given by

Let us find the percentage increase or decrease for the given cases.
a) 4x to x

Therefore, it is a percentage decrease (-75%) since it is negative.
b) 0.25m to 0.5m

Therefore, it is a percentage increase (100%) since it is positive.
c) 1/2p to 5/8p

Therefore, it is a percentage increase (25%) since it is positive
d) y to 0.68y

Therefore, it is a percentage decrease (-47%) since it is negative.