You are given
$\mathrm {p_1 = p}$
$\mathrm { p_n=k \cdot p_{n-1}+(1- k)(1-p_{n-1})}$
Write the latter in the form
$\eqalign{ & \mathrm {{p_n} = k{p_{n - 1}} + (1 - k)(1 - {p_{n - 1}})} \cr & \mathrm {{p_n} = k{p_{n - 1}} + 1 + k{p_{n - 1}} - k - {p_{n - 1}}} \cr & \mathrm {{p_n} = \left( {2k - 1} \right){p_{n - 1}} - (k - 1) }\cr} $
Now subtract $1/2$ from the equation and rearrange
$\eqalign{ & \mathrm {{p_n} = \left( {2k - 1} \right){p_{n - 1}} - \left( {k - 1} \right)} \cr &\mathrm { {p_n} - \frac{1}{2} = \left( {2k - 1} \right){p_{n - 1}} - \left( {k - 1} \right) - \frac{1}{2} } \cr & \mathrm {{p_n} - \frac{1}{2} = 2\left( {k - \frac{1}{2}} \right){p_{n - 1}} - \left( {k - \frac{1}{2}} \right) } \cr & \mathrm {{p_n} - \frac{1}{2} = \left( {2{p_{n - 1}} - 1} \right)\left( {k - \frac{1}{2}} \right) }\cr & \mathrm {{p_n} - \frac{1}{2} = \left( {{p_{n - 1}} - \frac{1}{2}} \right)\left( {2k - 1} \right) }\cr} $
Define now
$\mathrm {{p_n} - \frac{1}{2} = {u_n}}$
Then you have that, with $2 \mathrm k-1=\lambda$ ${\mathrm p_{\mathrm n}} - \frac{1}{2} = \left( {{ \mathrm p_{\mathrm n - 1}} - \frac{1}{2}} \right)\left( {2 \mathrm k - 1} \right)$ ${\mathrm u_n} = \lambda {\mathrm u_{\mathrm n - 1}}$
We get by induction on $\mathrm n$ that
${\mathrm u_{\mathrm n} = \mathrm \lambda ^{\mathrm n - 1} \mathrm u_1}$
Or that
$\eqalign{ & \mathrm {u_n} = {\lambda ^{\mathrm n - 1}}\mathrm {u_1} \cr & \mathrm {p_n} - \frac{1}{2} = {\left( {2 \mathrm k - 1} \right)^{\mathrm n - 1}}\left( \mathrm {p - \frac{1}{2}} \right) \cr & \mathrm {p_n} = {\left( {2 \mathrm k - 1} \right)^{\mathrm n - 1}}\left( \mathrm {p - \frac{1}{2}} \right) + \frac{1}{2} \cr} $