Junior Balkan MO

Solution:
We will first show that $PA$ is tangent to $(c)$ at $A$.
Since $E, D, Z, A$ are concyclic, then $\angle EDC = \angle EAZ = \angle EAB$. Since also the triangles $\triangle ABC$ and $\triangle EBA$ are similar, then $\angle EAB = \angle BCA$, therefore $\angle EDC = \angle BCA$.
Since $\angle FED = 90^\circ$, then $\angle PED = 90^\circ$ and so
$$\angle EPD = 90^\circ - \angle EDC = 90^\circ - \angle BCA = \angle EAC$$
Therefore the points $E, A, C, P$ are concyclic. It follows that $\angle CPA = 90^\circ$ and therefore the triangle $\triangle PAZ$ is right-angled. Since also $B$ is the midpoint of $AZ$, then $PB = AB = BZ$ and so $\angle ZPB = \angle PZB$.

Furthermore, $\angle EPD = \angle EAC = \angle CBA = \angle EBA$ from which it follows that the points $P, E, B, Z$ are also concyclic.
Now observe that
$$\angle PAE = \angle PCE = \angle ZPB - \angle PBE = \angle PZB - \angle PZE = \angle EZB$$
Therefore $PA$ is tangent to $(c)$ at $A$ as claimed.
It now follows that $TA = TZ$. Therefore
$$\begin{aligned} \angle PTZ & = 180^\circ - 2(\angle TAB) = 180^\circ - 2(\angle PAE + \angle EAB) = 180^\circ - 2(\angle ECP + \angle ACB) \\ & = 180^\circ - 2\left(90^\circ - \angle PZB\right) = 2(\angle PZB) = \angle PZB + \angle BPZ = \angle PBA . \end{aligned}$$
Thus $T, P, B, Z$ are concyclic, and since $P, E, B, Z$ are also concyclic then $T, E, B, Z$ are concyclic as required.