Hookes law problem - mechanics problem - CIE past papers

So today as i was practicing past papers, I came upon this question and honestly i coudnt get how they got the answer, here it is

A light elastic string has natural length 3m and modulus of elasticity 45N. A particle P of weight 6N is attached to the mid-point of the string. The ends of the string are attached to fixed points A and B which lie in the same vertical line with A above B and AB = 4. The particle P is released from rest at the point 1.5m vertically below A.
Calculate the distance P moves after its release before first coming to instantaneous rest at a point vertically above B. (you may assume that at this point the part of the string joining P to B is slack).

I'd reaaally appreciate your help guys.

What have you tried? At what point do you get stuck? Post your working, please.

l=3, (lmabda)= 45N, PB = ?

45 * (3-1.5-PB)^2 / 2*3 = 6 (PB)
45 * (2.25 + PB^2 - 3PB)/6 = 6PB
7.5 * (2.25 + PB ^ 2 - 3PB) = 6PB
PB ^ 2 - 3.8PB + 2.25 = 0
= 3.06 , 0.735

I attempted to equate PE with EE since PE is decreasing (moving downwards) and EE is increasing (tension is increasing in the part AP of the string), but the correct answer was 1.22m.