Let the segments $T E$ and $T F$ cross $A C$ at $P$ and $Q$, respectively. Since $P E \| C D$ and $E D$ is tangent to the circumcircle of $A B C D$, we have
$$\angle E P A=\angle D C A=\angle E D A,$$
and so the points $A, P, D$, and $E$ lie on some circle $\alpha$. Similarly, the points $C, Q, D$, and $F$ lie on some circle $\gamma$. We now want to prove that the line $D T$ is tangent to both $\alpha$ and $\gamma$ at $D$. Indeed, since $\angle F C D + \angle E A D = 180^{\circ}$, the circles $\alpha$ and $\gamma$ are tangent to each other at $D$. To prove that $T$ lies on their common tangent line at $D$ (i.e., on their radical axis), it suffices to check that $T P \cdot T E = T Q \cdot T F$, or that the quadrilateral $P E F Q$ is cyclic. This fact follows from
$$\angle Q F E = \angle A D E = \angle A P E.$$
Since $T D = T K$, we have $\angle T K D = \angle T D K$. Next, as $T D$ and $D E$ are tangent to $\alpha$ and $\Omega$, respectively, we obtain
$$\angle T K D = \angle T D K = \angle E A D = \angle B D E,$$
which implies $T K \| B D$. Next we prove that the five points $T, P, Q, D$, and $K$ lie on some circle $\tau$. Indeed, since $T D$ is tangent to the circle $\alpha$ we have
$$\angle E P D = \angle T D F = \angle T K D,$$
which means that the point $P$ lies on the circle (TDK). Similarly, we have $Q \in (T D K)$. Finally, we prove that $P K \| B C$. Indeed, using the circles $\tau$ and $\gamma$ we conclude that
$$\angle P K D = \angle P Q D = \angle D F C,$$
which means that $P K \| B C$. Triangles $T P K$ and $D C B$ have pairwise parallel sides, which implies the fact that $T D, P C$ and $K B$ are concurrent, as desired. !

Comment 1. There are several variations of the above solution. E.g., after finding circles $\alpha$ and $\gamma$, one can notice that there exists a homothety $h$ mapping the triangle $T P Q$ to the triangle $D C A$; the centre of that homothety is $Y = A C \cap T D$. Since
$$\angle D P E = \angle D A E = \angle D C B = \angle D Q T,$$
the quadrilateral $T P D Q$ is inscribed in some circle $\tau$. We have $h(\tau) = \Omega$, so the point $D^{*} = h(D)$ lies on $\Omega$.
Finally, by
$$\angle D C D^{*} = \angle T P D = \angle B A D,$$
the points $B$ and $D^{*}$ are symmetric with respect to the diameter of $\Omega$ passing through $D$. This yields $D B = D D^{*}$ and $B D^{*} \| E F$, so $h(K) = B$, and $B K$ passes through $Y$.