Construct the segment $AK$. By condition, $T$ lies on $AK$. Let $R$ be the intersection point of lines $DT$ and $KM$ (see the figures). Let $AN = x$, $ND = y$, $RM = z$, $AT : TK = 1 : k$. By Thales' theorem, $k = TK : AT = y : x$ (since $NT \parallel DK$). Since $NMKD$ is a parallelogram, we have $MK = ND = y$, and then
$$z : y = RM : MK = [TM \parallel DK] = RT : TD = [TC \parallel RK] = KC : CD =$$
$$= [TC \parallel AD] = KT : TA = k. \quad (1)$$
So, $y = kx$, $z = ky = k^2x$. Let $P_1$ be the point of intersection of $AM$ and $RD$ (Fig. 1).

Construct a line passing through $A$ parallel to $AM$, and let $Q_1$ be the point of intersection of this line and the line $MK$. By Thales' theorem,
$$\begin{gathered} RP_1 : P_1D = RM : MQ_1 = [AMQ_1D \text{ is a parallelogram,} \\ MQ_1 = AD = x + y] = z : (x + y) = k^2x : (x + kx) = \frac{k^2}{1+k}. \end{gathered}$$
Therefore,
$$RP_1 = \frac{k^2}{1+k} P_1D = \frac{k^2}{1+k} (RD - RP_1) \Rightarrow RP_1 = \frac{k^2}{1+k+k^2} RD. \quad (2)$$

Let $P_2$ be the point of intersection of the segments $BK$ and $RD$ (Fig. 2). Construct a line passing through $T$ parallel to $BK$, and let $Q_2$ be the point of intersection of this line with the line $MK$. By Thales' theorem,
$$RP_2 : P_2T = RK : KQ_2 = [BKQ_2T, BTNA, \text{NMKD is a parallelogram,}]$$
$$KQ_2 = BT = AN = x, MK = ND = y] = (z + y) : x = (k^2 + k).$$
Therefore,
$$RP_2 = (k^2 + k)P_2T = k(k + 1)(RT - RP_2) \implies$$
$$RP_2 = \frac{k(k+1)}{k^2+k+1}RT = [RT = kTD = k(RD - RT) \implies$$
$$RT = \frac{k}{k+1}RD] = \frac{k^2}{k^2+k+1}RD.$$
Comparing this result with (2), we see that $P_1$ coincides with $P_2$, and so they coincide with $P$, which gives the required statement.