Let $BC$ touch $k$ in $T_k$ and $DE$ touch $l$ in $T_l$. We shall prove the required fixed point is $T_k$.

First we show the points $T_k$, $T_l$ and $P$ are collinear. Denote by $U$ and $V$ the points where $EP$ and $DP$ touch $k$, and $M$ and $N$ the points where $EP$ and $DP$ intersect $BC$. Let $T_1$ and $T_2$ be the tangent points of $k$ and the rays $AB$ and $AC$ respectively.

Fig. 1

As $BC \parallel DE$, the triangle $DEP$ is similar to $NMP$ and the homothety $H$ with the centre $P$ and the quotient $q = MN/ED$ maps the segment $DE$ into $NM$. To prove the collinearity of $T_k, T_l, P$, it suffices to derive the equality
$$\frac{MT_k}{NT_k} = \frac{ET_l}{DT_l} \qquad (1)$$
if this is true, $H$ maps $T_l$ into $T_k$.

Let $a, b, c$ be the lengths of the sides in the triangle $DEP$ as in fig. 1. Set $AD = c$, $AE = d$. Let us remind the well-known formulae for the length of the segments between the vertex of a triangle and the tangent points of its incircle and excircle: In any triangle $XYZ$, the distance of $X$ from the tangent point of the incircle and excircle (lying on $XY$) is $(XY + XZ - YZ)/2$ and $(XY + YZ - XZ)/2$ respectively.

The circle $k$ is the excircle of $NMP$. Hence
$$\frac{MT_k}{NT_k} = \frac{(MN + NP - MP)/2}{(MN + MP - NP)/2} = \frac{qa + qc - qb}{qa + qb - qc} = \frac{a+c-b}{a+b-c} \qquad (2)$$
The circle $l$ is the incircle of $DEA$. Hence
$$\frac{ET_l}{DT_l} = \frac{(DE + AE - AD)/2}{(DE + AD - AE)/2} = \frac{a+d-c}{a+c-d} \qquad (3)$$
When we draw two tangent lines from a point to a circle, the distances of the two tangent points from the original point are equal. Repeating this argument several times we get
$$c + c + PU = e + c + PV = e + DT_l \quad AT_1 \cdot AT_2 = d + ET_2 \cdot d + b + PU$$
so $e + c = d + b$. Then $c - b = d - e$ and substituting into (2) and (3) gives
$$\frac{MT_k}{NT_k} = \frac{a + (c - b)}{a - (c - b)} = \frac{a + (d - e)}{a - (d - e)} = \frac{ET_l}{DT_l}$$
which is exactly (1). Hence $P$ lies on $T_1T_k$.

Similarly we show that also $T_k$, $T_l$ and $Q$ are collinear. Denote by $U'$ and $V'$ the points where $CQ$ and $BQ$ touch $l$, and $M'$ and $N'$ the points where $C'Q$ and $B'Q$ intersect $DE$. Let $T_1'$ and $T_2'$ be the tangent points of $l$ and the rays $AD$ and $AE$ respectively. Let $a', b', c'$ be the lengths of the sides in the triangle $BCQ$ and $AB = c'$, $AC = d'$.

Fig. 2

Repeating the arguments from the first part (here both circles are excircles) we get
$$\frac{M'T_l}{N'T_l} = \frac{a' + c' - b'}{a' + b' - c'} \qquad \frac{CT_k}{BT_k} = \frac{a' + c' - d'}{a' + d' - c'}$$
Comparing the lengths (fig. 2) gives
$$c' - c' - QU' = c' - c' - QV' = c' - BT_{1}' = AT_{1}' = AT_{2}' = d' - CT_{2}' = d' - b' - QU'$$
so $c' - b' = c' - d'$ and
$$\frac{M'T_l}{N'T_l} = \frac{CT_k}{BT_k}$$
Finally, using the homothety of $BCQ$ and $N'M'Q$ we conclude that $Q$ lies on $T_lT_k$.

Therefore the line $PQ$ (obviously, $P \neq Q$) is identical to the line $T_lT_k$ and passes through $T_k$, which is independent of the choice of $p$.