Denote $CD \cap {\omega _1} = G$, then $\angle GDF = \angle CDM = 90^\circ $, so $FG$ is the diameter of ${\omega _1}$, hence $P$ is the homothetic center of ${\omega _1}$ and $\omega $, so the centers of ${\omega _1}$ and $\omega $ are all in the line $PZ$. Because $D$ is the homothetic center of ${\omega _1}$ and $\Omega $, so $M,D,F$ are collinear. Because $\angle CEF = 90^\circ $, $ \angle FDC=\angle MDC =90^\circ $, $D,F,E,C$ are concyclic, so $MD \cdot MF = ME \cdot MC$, $MZ$ is the radical line of ${\omega _1}$ and $\omega $, so $MZ \bot ZP$, hence $\angle CZM = \angle EZP$, and we have $\Delta CZM \sim \Delta EZP$, so $\tan \angle ZEP = \tan \angle ECZ = \frac{{EZ}}{{CZ}} = \frac{{PE}}{{MC}}$.

Here my solution: Obviously, $ P $ is exsimilicenter of $ w $ and $ w_1 $, hence $ P,O,O_1,Z $ are collinear, where $ O ,O_1 $ be the centers of $ w,w_1 $, respectively.We have easily that $ \angle PEZ=\angle ZCE $, and $ \angle MZC=\angle PZE $ (because we know that $ MZ $ is common internal tangent of $ w $ and $ w_1 $ ).Thus the triangles $ PZE $ and $ CZM $ are relatively similar.Hence $ tan \angle ZEP=tan \angle ECZ= \frac{ZE}{CZ}=\frac{PE}{CM} $, so we are done !

Homothety that takes $w$ to $\omega$ takes to midpoint of arc $AB$ so we have $C-E-M$ so $CE$ is perpendicular on $AB$ and analogous $D-F-M$.Trivialy $\tan \angle ZEF=\frac{ZE}{ZC}$ so for $\triangle EZP \sim \triangle CZM$ we only need $MZ $ perpendicular $PZ$ and we have that $\angle ZPE=\angle ZCM$, now Mongue de Alambert theorem we have $C$ as center homothety $w$ to $\omega$, $D$ as center of homothety $w_{2}$ to $\omega$ so center of homothety $w$ to $w_{1}$ lies on $CD$,so because $FE$ is common tangent we have $P-O_{1}-Z-O_{2}$.Now we need to prove that $MZ$ is common tangent of $w_1$, $w_{2}$ which is true because $CDEF$ is cyclic so $M$ is on radical axis which is common tangent.

Denote $O_\Omega$, $O_w$ and $O_{w_1}$ as the centers of circles $\Omega$, $w$ and $w_1$. It is well known that the power of $M$, the midpoint of major arc $BC$, with respect to a circle tangent to $BC$ and minor arc $BC$ is equal to $MA^2=MB^2$. It follows that M is on the radical axis of $O_{w_1}$ and $O_w$, implying that $MZ \perp O_{w_1}O_w$. Since $\angle{PEM}=90^{\circ}$, we have that quadrilateral $PZEM$ is cyclic. This means that $\angle{ZPE}=\angle{ZME}$, and since $\angle{ZEP}=\angle{ZCE}$, we have that $\triangle{PZE}$ is similar to $\triangle{ZMC}$, implying that $\frac{PE}{ZE}=\frac{MC}{ZC}$. Rearranging we have that $\frac{PE}{CM}=\frac{ZE}{ZC}$. Since $\angle{CZE}=90^{\circ}$, we have that $\tan{\angle{ZCE}}=\frac{ZE}{ZC}=\frac{PE}{CM}$. However, since $\angle{ZCE}=\angle{ZEP}$, we obtain $\tan{\angle{ZEP}}=\frac{PE}{CM}$ as desired. $\square$

Computational Solution:

Dear yayups and MLs, Very nice solution, in the style of the Great Yetti. Happy New Year to all ! M.T.

A question, are $C,Z,F$ collinear? PS: Okay, never mind, I got my doubt clarified myself.

Since $\tan \angle ZEP = \tan \angle ZCE = \tfrac{ZE}{ZC}$, we will prove that $\tfrac{ZE}{ZC}=\tfrac{PE}{CM}$ by showing that $\triangle{ZEP}\sim\triangle{ZCM}$. By a homothety at $D$ sending $\omega_1$ to $(ABC)$, we know that $D$, $F$, and $M$ are collinear. In addition, $ME\cdot MC=MB^2=MF\cdot MD$ by considering similar triangles $MEB/MBC$ and $MFB/MBD$. Hence, $M$ is on the radical axis of $\omega$ and $\omega_1$. This also means that $\triangle{ZCM}\sim\triangle{EZM}$. By Monge's Theorem with circles $(ABC)$, $\omega$, and $\omega_1$, $C$, $D$, and $P'$ are collinear, where $P'$ is intersection of the common external tangents of $\omega$ and $\omega_1$. Since $P'$ is on $\overleftrightarrow{CD}$ and $\overleftrightarrow{EF}$, we must have $P'=P$. Then $\angle PZM = 90^{\circ}=\angle PEM$, so $PZEM$ is cyclic. We have that $\angle{EMZ}=\angle{EPZ}$ and $\angle{ZEP}=\angle{EZM}=\angle{ECZ}$, so $\triangle{EZM}\sim\triangle{ZEP}$ (this is actually a congruence). Therefore, $\triangle{ZEP}\sim\triangle{EZM}\sim\triangle{ZCM}$ and $\tan \angle ZEP = \tan \angle ZCE = \tfrac{ZE}{ZC}=\tfrac{PE}{CM}$, as desired.

Let $O_2$ be the center of $\omega$ and $O_3$ be the center of $\omega_1$. We first note that because $E$ is the midpoint of $\overline{AB}$ and $M$ is the midpoint of arc $\overline{AB}$, we have that $\angle MEB = 90^\circ$. By the properties of circles inscribed within segments, $M$, $E$, and $C$ are collinear, so $\angle CEB = 90^\circ$. Since $\angle CEB = 90^\circ$ and $\omega$ is tangent to $AB$ at $E$, $CE$ is a diameter of $\omega$. Since $\angle ZEP = \angle ZCM$, it suffices to show that $\angle EPZ = \angle CMZ$. Since $\omega$ and $\omega_1$ intersect at just $Z$, their radical axis is the line going through $Z$ that is tangent to both $\omega$ and $\omega_1$. By the properties of circles inscribed in segments, $\mbox{Pow}_{\omega_1}(M) = MB^2 = \mbox{Pow}_{\omega}(M)$. Thus, $M$ is on the radical axis of $\omega$ and $\omega_1$ and $MZ$ is tangent to both $\omega$ and $\omega_1$. Let the other intersection of $PC$ with $\omega$ be $C'$. This intersection exists because we would otherwise have that $PC$ is perpendicular to a diameter of $\omega$, $CE$, while $PE$ is also perpendicular to $CE$, making $P$ not exist on the Euclidean plane. By Power of a Point, $PC' \cdot PC = PE^2$, meaning that $\frac{PE}{PC} = \frac{PC'}{PE}$ and $\triangle PEC \sim \triangle PC'E$. Since $\angle CEP = \angle CEB = 90^\circ$, we then have that $\angle EC'P = 90^\circ$. Since $MC$ is a diameter of $\Omega$, $\angle MDC = 90^\circ$. By the properties of circles inscribed in a segment, $M$, $D$, and $F$ are collinear, so $\angle FDC = 90^\circ$. Now, consider the homothety about $P$ taking $E$ to $F$, $\mathcal{H}$. Since $\angle FDP = 90^\circ = \angle EC'P$, $\mathcal{H}(C') = D$. Since $\omega$ goes through $C'$ and is tangent to $EP$ at $E$ and $\omega_1$ goes through $D$ and is tangent to $FP$ at $F$, we must have that $\mathcal{H}(\omega) = \omega_1$. Thus, we then have that $\mathcal{H}(O_2) = O_3$, meaning that $O_2$, $O_3$, and $P$ are collinear. Since $\omega$ and $\omega_1$ are tangent to each other at $Z$, we have that $O_2$, $Z$, and $O_3$ are collinear. Thus, $P$, $O_3$, and $Z$ are collinear. Since $O_3Z$ is perpendicular to $MZ$, we have that $PZ$ is perpendicular to $MZ$, and so $\angle PZM = 90^\circ$. Since $\angle PZM = 90^\circ$ and $\angle PEM = \angle MEP = \angle MEB = 90^\circ$, we have that $PZEM$ is a cyclic quadrilateral. Now, we have that $\angle EPZ = \angle EMZ = \angle CMZ$. Because $\angle ZEP = \angle ZCM$, we have that $\triangle ZEP \sim \triangle ZCM$. Then, we have that $\frac{EZ}{CZ} = \frac{PE}{CM}$. Since $\angle CZE = 90^\circ$, we have that $\frac{EZ}{CZ} = \tan \angle ZCE = \tan \angle ZEP$. Thus, we have that $\tan \angle ZEP = \frac{PE}{CM}$, and we are done. $\fbox{}$

Let $O_1$ and $O_2$ be the centers of $\omega$ and $\omega_1$. First of all note that by Archimedes + Shooting Lemma we have $CE \cap DF = M$ and $$MA^2 = MB^2 = ME \cdot MC = MF \cdot MD$$so $EFDC$ is cyclic. Claim: $O_1,Z,O_2,P$ are collinear. Proof: Let $E'$ and $F'$ be the reflections of $E$ and $F$ across $O_1ZO_2$ (which we know to be collinear). By PoP we have$$PF \cdot PE = PF' \cdot PE' =PD \cdot PC$$hence $E'F'DC$ is cyclic. By definition we know $EFF'E'$ is isosceles trapezoid so it is cyclic. We use radical axis theorem on the three circles $(EFF'E'), (EFDC), (E'F'DC)$ to obtain $E'F', EF, CD$ concur at $P$. Obviously $E'F', O_1O_2, EF$ must concur so $O_1O_2$ passes through $P$ as desired. $\square$ Since $E$ is the midpoint of $AB$ we know $CE$ is the diameter of $\omega$. Since $AB$ tangent to $\omega$, we know$$\angle ZEP = \angle ZCM$$and since $CEDF$ is cyclic, $M$ lies on the tangent at $Z$. Therefore, by angle chasing with $MZ$ tangent to $\omega$, we get$$\angle CZM = 180^{\circ} - \angle CEZ = 180^{\circ} - \angle O_1ZE = \angle EZP$$since $O_1, Z, P$ collinear from our claim. Therefore, by AA similarity we have $\triangle CZM \sim \triangle EZP$ so therefore$$\tan{\angle ZEP} = \tan{\angle ZCM} = \frac{EZ}{ZC} = \frac{PE}{CM}$$as desired. $\blacksquare$ reeeeeeee

I have used Yayups' diagram

Claim 1 : $\overline {C - E - M}$ and $\overline {D - F - M}$ Proof : This follows from Shooting Lemma. Claim 2 : $CEFD$ is a cyclic quadrilateral . Proof : $MZ$ is the radical axis of $\omega_1$ and $\omega$. Now it is just Radical Axis Theorem. From Claim 2 $\angle CEP = \angle CDF = 90^{\circ} \Longrightarrow CM$ is the diameter of $\Omega$ and $FG$ is the diameter of $\omega_1$. So, there is a homothety centred at $P$ mapping $\omega_1 \mapsto \omega \Longrightarrow \overline{P - Z - T}$ Thus, $\angle PZM = 90^{\circ} \Longrightarrow PEZ\sim MCZ \Longrightarrow \tan \angle ZEP = \frac{EZ}{ZC} = \frac{PE}{CM}$

Note that $\angle ZEP=\angle ZCE$ and $PE=CE\tan\angle DCE$. As $\angle MDC=\frac{\pi}{2}$, we have $\tan\angle DCE=\frac{MD}{DC}$. Hence $PE=CE\frac{MD}{DC}$. Now consider an inversion $f$ with centre at $C$ and radius 1. The resulting diagram is attached. Clearly, $|f(Z)f(E)|=|f(D)f(M)|$. Hence $\tan\angle ZCE=\tan\angle f(E)Cf(Z)=\frac{|f(Z)f(E)|}{|f(E)C|}=\frac{|f(D)f(M)|}{|f(E)f(M)|+|f(M)C|}=\frac{\frac{MD}{CM\cdot CD}}{\frac{EM}{CE\cdot CM}+\frac{1}{CM}}=\frac{\frac{MD}{DC}}{\frac{CM}{CE}}=\frac{\frac{PE}{CE}}{\frac{CM}{CE}}=\frac{PE}{CM}$. QED

Let $O_1$ be the center of $\Omega$, $O_2$ be the center of $\omega$, and $O_3$ be the center of $\omega_1$. Also, let $G$ be the point on $\omega$ diametrically opposite to $F$. Lemma: $O_2, Z, O_3, P$ are collinear, $C, Z, F$ are collinear, $E, Z, G$ are collinear, and $D, F, M$ are collinear. Proof: It is well-known that $D, F, M$ are collinear (quick sketch of the proof: the homothety about $D$ sending $\omega_1$ to $\Omega$ sends $F$ to $M$). $C, Z, F$ are collinear and $E, Z, G$ are collinear because the negative homothety about $Z$ sending $O_2$ to $O_3$ sends $C$ to $F$ and $e$ to $G$. It is obvious that $O_2, O_3, Z$ are collinear, and the homothety about $P$ sending $\triangle{PFG}$ to $\triangle{PEC}$ sends $O_2$ to $O_3$ (since they are midpoints of $GF, CE$ respectively), so $P, O_2, O_3$ are collinear, and we have proven all four parts of the lemma. End lemma. Now note that $\angle{ZCE}=\angle{ZEP}$, so it suffices to show that $\tan{\angle{ZCE}}=\frac{PE}{CM}$. But $\tan{\angle{ZCE}}=\frac{EZ}{CZ}$, so thus it suffices to prove that $\frac{EZ}{PE}=\frac{CZ}{CM}$. But note since $\angle{ZCE}=\angle{ZEP}$, this is equivalent to proving $\triangle{PEZ}~\triangle{MCZ}$, which in turn is equivalent to proving $\angle{CMZ}=\angle{EPZ}$. But note that we have $\angle{ECZ}=\angle{O_2ZC}=\angle{PZF}$, and we also have $\angle{ECZ}=\angle{ZEP}=\angle{EZM}$, meaning that $\angle{EZM}=\angle{PZF}$, so thus $\angle{PZM}=90$ because $\angle{EZF}=90$. But since $\angle{PEM}=90$, we have $PZEM$ is cyclic, so thus $\angle{EMZ}=\angle{CMZ}=\angle{EPZ}$, and hence we are done.

It's well known $DFM$ collinear, and it's clear this line is perpendicular to $CP$, hence $F$ is the orthocenter of $\triangle CMP$. Considering the homothety at $Z$ sending $\omega_1$ to $\omega_2$ we see $CZF$ are collinear, and this line is perpendicular to $PM$ by our previous observations. Since $CEDF$ is cyclic by radical axis we have $MZ$ is tangent to $\omega_1,\omega_2$, so $\angle EZM = \angle ZEP$. In addition, $EZ \perp CZ \perp MP$ so $EZPM$ is an isosceles trapezoid. To finish, we have $$\tan(\angle ZEP) = \tan(\angle ECF) = \tan(90^\circ - \angle CMP) = \frac{EM}{EP}$$so it suffices to show $\frac{EM}{EP}=\frac{PE}{CM}$, which follows since $EM \cdot CM = MZ^2 = PE^2$.

Quite nice, although it's a bit tricky to pick which length ratio to focus on (I originally tried to do $\frac{PF}{PE}=\frac{EM}{CM}$ which failed). Needed a few hints to solve. Since $\angle CEF=\angle EZC=90^\circ$, we have $\triangle CZF \sim \triangle CEF \sim \triangle EZF$, so $\tan\angle ZEP=\tan\angle ZEF=\tan\angle ECZ=\frac{EZ}{CZ}$. Hence we just have to prove: $$\frac{EZ}{CZ}=\frac{PE}{MC}.$$Now let $G \neq D$ denote the point where $\overline{CP}$ intersects $\omega_1$. We then have $\angle GDF=\angle CDM=90^\circ$, hence $\overline{FG}$ is a diameter. This implies that the homothety sending $\omega_1$ to $\omega$ is centered at $P$. Observe that the homothety sending $\omega$ to $\Omega$ sends $E$ to $M$ and is centered at $C$, hence $C,E,M$ are collinear. Similarly, $D,F,M$ are collinear. Also, as $\angle CEF=\angle CDF=90^\circ$, we have $C,D,E,F$ concyclic. Thus $ME\cdot MC=MF \cdot MD$, so $M$ lies on the radical axis of $\omega$ and $\omega_1$. As the perpendicular to the radical axis (passing through the centers) also passes through $P$, we obtain $\overline{PZ} \perp \overline{ZM}$. Since $\angle CZE=90^\circ$, we have $\angle CZM=\angle PZE$. We can also obtain that $\angle PEZ=\angle MCZ$ from the fact that $\triangle CEF$ is right with $Z$ as the foot of the $E$-altitude, hence $\triangle PZE \sim \triangle MZC$, from which we obtain $\frac{EZ}{CZ}=\frac{PE}{MC}$ as desired. $\blacksquare$

i am definitely not sleep-deprived while writing this. also had the same issue as @above, focused on the wrong ratio at first Since $\angle ZEP=\angle ZCE$, it suffices to prove \begin{align*} \frac{PE}{CM}=\frac{ZE}{CM}\implies\frac{PE}{ZE}=\frac{CM}{CZ}, \end{align*}or $\triangle PEZ\sim\triangle MCZ$. Manipulating ratios, we have \begin{align*} \frac{PZ}{ZE}=\frac{MZ}{CZ}\implies\frac{PZ}{MZ}=\frac{ZE}{CZ}, \end{align*}which means it is the same to prove $\triangle MZP\sim\triangle CZE$. However, this is also the same as proving $MEZP$ is cyclic, as if so, then $\angle MZP=\angle MEP=90^\circ=\angle CZE$ and $\angle ZEP=\angle ZEB=\angle ECZ$. Moreover, proving that $MZ$ is tangent to $\omega$ finishes, as if so, $\angle MZP=90^\circ=\angle MEP$ which implies the cyclicity as needed. Indeed, note that $M$ is on the radical axis of $\omega$ and $\omega_1$ since $MA^2=ME\cdot MC=MF\cdot MD$ by Shooting Lemma. $Z$ is trivially on the radical axis as well, and since $\omega$ and $\omega_1$ are externally tangent so we may conclude.

$C,E,M$ collinear $D,F,M$ collinear $C,Z,F$ collinear Remaining sinus Law

It is easy to see that $\angle ZEP \cong \angle ZCM$. Then $\tan \angle ZEP = \tan \angle ZCM = \tan \angle ZCE = \frac{EZ}{ZC}$. We need to show this is the same as $\frac{PE}{CM}$. Notice that the power of $M$ w.r.t to $\omega_1, \omega_2$ is $MA^2=MB^2$, so $M$ lies on the radical axis, along with $Z$. Then $\angle MZO_1 = 90$, so $\angle CMZ = \angle O_1 MZ = 90 - \angle ZO_1 M = 90 - \angle PO_1 E = \angle O_1 P E = \angle ZPE$. Then by AA, $\triangle EZP \sim \triangle CZM$, so $\frac{EZ}{ZC} = \frac{PE}{CM}$, as desired.

Let $O$ and $O_1$ be the centers of $\omega$ and $\omega_1$ respectively, and let $X$ be the intersection of $PC$ with $\omega_1$ with $X\neq D$ and $Y$ be the foot of the perpendicular from $X$ to $EC$. $C, E, M$ are collinear and it's well known that $D, F, M$ are collinear. $C, Z, F$ are collinear because $\angle{O_1ZF}=\angle{O_1FZ}=\angle{ZCO}=\angle{OZC}$ because $FX\parallel EC$. Since $\triangle{ZEF}\sim\triangle{ECF}$, $\angle{ZEP}=\angle{ECF}$. Call $H$ the orthocenter of $\triangle{CEX}$. Because $\triangle{CHX}\sim\triangle{CFP}$ and $\triangle{CEH}\sim\triangle{CMF}$, we get $\frac{CX}{CP}=\frac{CH}{CF}=\frac{CE}{CM}$ which means $\triangle{CEX}\sim\triangle{CMP}$. Hence, $\tan{\angle{ZEP}}=\tan{\angle{ECF}}=\frac{FE}{CE}=\frac{XY}{CE}=\frac{PE}{CM}$, as desired.

Sardor wrote: Here my solution: Obviously, $ P $ is exsimilicenter of $ w $ and $ w_1 $, hence $ P,O,O_1,Z $ are collinear, where $ O ,O_1 $ be the centers of $ w,w_1 $, respectively.We have easily that $ \angle PEZ=\angle ZCE $, and $ \angle MZC=\angle PZE $ (because we know that $ MZ $ is common internal tangent of $ w $ and $ w_1 $ ).Thus the triangles $ PZE $ and $ CZM $ are relatively similar.Hence $ tan \angle ZEP=tan \angle ECZ= \frac{ZE}{CZ}=\frac{PE}{CM} $, so we are done ! Why $P, O_1, Z$ and $P$ are collinear?

@above There is a homothety that takes $\omega _1$ to $\omega$. Lilavati's solution explains it well.

Clarly $C-E-M$ by shooting lemma, so as $E$ is the midpoint of $AB$ we get that $CM$ is diameter of $\Omega$ and $CE$ is diameter of $\omega.$ Now notice that $\angle ZEP=\angle ZCE,$ and $\tan \angle ZCE=\frac{ZC}{CE}.$ We will show that $\triangle ZEP\sim \triangle ZCM,$ which will trivially imply that we are done. As $\angle ZEP=\angle ZCE$ we just need to show $\angle PZE=\angle MZC:$ For this invert at $M$ with radius $MA=MB.$ Clarly $\omega \to \omega, AB\to \Omega, \Omega\to AB.$ So this shows that $(DZF)\to(DZF),$ implying that $MZ$ is tangent to both $\omega, (DZF).$ We now claim that $\angle PZM=90^\circ.$ Notice that by the tangency, to show this is equivalent to showing that $PZ$ passes through the centers of $(DZF)$ and $\omega.$ But this is true by Monge d'Alembert on $(DZF), \omega, \Omega,$ as this shows that $P$ is the exscimillicenter of $(DZF), \omega.$ Now just end the problem by computing $\angle PZE=90^\circ+\angle MZE=\angle MZC,$ solving the problem.