Here is my solution without any random equilateral triangles. Let's use phantom points instead. Let $P'$ be the intersection point of $ME$ and $AC$. Let $AC$ and $BD$ intersect at point $O$. Obviously $\bigtriangleup DBF \cong \bigtriangleup MEF$ ($DF=FM, FE=BF, \angle DFB=60^{\circ}+\angle MFB=\angle MFE$). Hence $\angle DBF=\angle MEF$. Let $DB$ intersects $ME$ at point $K$. We have $\angle KMF=\angle EMF=\angle BDF=\angle KDF.$ So $MKFD$ is cyclic. Hence $\angle MFK=\angle MDK=\angle ADB=45^{\circ}$. We have $OM=AM=MF$ and $\angle KOM=\angle DOM=45^{\circ}$ because $ABCD$ is a square. So $\bigtriangleup MOF$ is isosceles. We have $\angle MFK=\angle KOM=45^{\circ}$ and $\angle MFO=\angle MOF$. So $\angle KFO=\angle KOF$ and hence $FK$=$OK$. Therefore $MK$ is the perpendicular bisector of $OF$. Hence $MP'$ is the angle bisector of $\angle AP'F$, so we only have to prove that $P\equiv P'$ and we will be ready. We have $\angle KMO=\angle KMF=15^{\circ}=\angle CAF=\angle P’AF$. Thus $AP’FM$ is cyclic. So $\angle OP'F=2\angle MP'F=2\angle MAF=60^{\circ}$. Also $FP'=P'O$ so $\angle OFP'=60^{\circ}$ and $FO=FP'$. We have $\bigtriangleup DFO \cong \bigtriangleup MFP'$ because $DF=MF, FO=FP', \angle DFO=60^{\circ}+\angle MFO=\angle MFP'$. Hence $\frac{ME}{2}=\frac{BD}{2}=OD=MP'$. Therefore $P'$ is the midpoint of $ME$. Hence $P'\equiv P$, as desired.

a) Let $O$ be center of $ABCD$. We have $MF = DF, FE = FB$ and $\angle{MFE} = \angle{MFB} + \angle{BFE} = \angle{MFB} + \angle{MED} = \angle{DFB}$. Then $\triangle MFE = \triangle DFB$. But $O, P$ are midpoints of $BD, EM$ so $\triangle BFP = \triangle DFO$. Hence $\angle{MPF} = \angle{DOF} = \angle{DAF} = 30^{\circ}$. From this, we have $A, M, F, P$ lie on a circle. Therefore, $\angle{MAP} = 180^{\circ} - \angle{MFP} = 180^{\circ} - \angle{DFO} = \angle{DAO}$ or $P \in AC$ b) Since $MA = MD = MP$ and $A, M, F, P$ lie on a circle, we have $PM$ is bisector of $\angle{APF}$