We drop perpendiculars from P which is the intersection of extensions of HN and FM on EF and HK. J and I are the foot of these perpendiculars respectively. We have:

$PF\bot BL , and, PJ\bot JF\Rightarrow \angle JPF=\angle BLF$

So right angled triangles PFJ and LFB are congruent due to ASA and we have:

$PJ=LF=\frac{EA}2$

Similarly triangles HPI and DHL are congruent and we have:

$PI=\frac {AK}2$

Also :

$AK=EA$

That is P has equal distance from EL and LK, so it is coincident on the intersection of diagonals of the square ELKA.

We drop perpendiculars from P which is the intersection of extensions of HN and FM on EF and HK. J and I are the foot of these perpendiculars respectively. We have:

$PF\bot BL , and, PJ\bot JF\Rightarrow \angle JPF=\angle BLF$

Also:

$\triangle LFB=\triangle PFG\Rightarrow PF=BL$

So right angled triangles PFJ and LFB are congruent due to ASA and we have:

$PJ=LF=\frac{EA}2$

Similarly triangles HPI and DHL are congruent and we have:

$PI=\frac {AK}2$

Also :

$AK=EA$

That is P has equal distance from EL and LK, so it is coincident on the intersection of diagonals of the square ELKA.