Here we know that parallelogram ABCD and rectangle ABEF are on the same base AB and between the same parallels such that:

AB = CD and AB = EF

So, CD = FE

Now, adding AB on both sides

AB + CD = AB + FE (i)

Since we know that hypotenuse is the longest side of a triangle

∴ AD > AF (ii)

And, BC > BE (iii)

Adding (ii) and (iii),

​ AD + BC > AF + BE (iv)

Now, Perimeter of ABCD = AB + BC + CD + AD

And, Perimeter of ABEF = AB + BE + FE + AF

​Adding (i) and (iv),

AB + CD + AD + BC > AB + FE + AF + BE

Thus, we can say that the perimeter of parallelogram ABCD is greater than that of rectangle ABEF.