Probability distribution for y=min(x,a) let's say a random x has the following pdf, then how we can find the distribution for $y=min (x,a)$, for $\theta <= a <=w $?
$(x-\theta)^k exp(-\lambda(x-\theta))$. for $\theta<=x<=w$
Thank you for helping!

$\theta$ is just a shift parameter, either for $x$ and $a$ and $w$ and thus also for $y$,  and we can get rid of it, leaving
$$
\left\{ \begin{gathered}
  0 \leqslant a \leqslant w\quad p(a) = 1/w\quad \left( {\text{?}\;\text{supposed}} \right) \hfill \\
  0 \leqslant x \leqslant w\quad p(x) = x^{\,k} e^{\, - \,\lambda \,x} /\int_{x = 0}^{\,w} {x^{\,k} e^{\, - \,\lambda \,x} dx}  = x^{\,k} e^{\, - \,\lambda \,x} /C\quad \left( {\text{?}\;\text{supposed}} \right) \hfill \\
  0 \leqslant y = \min (a,x) \leqslant w \hfill \\ 
\end{gathered}  \right.
$$
where it is understood that all the parameters are net of $\theta$.  
Then
$$
\begin{gathered}
  p(y)dy = P\left( {y \leqslant a \leqslant y + dy} \right)P\left( {y \leqslant x \leqslant w} \right) + P\left( {y \leqslant x \leqslant y + dy} \right)P\left( {y \leqslant a \leqslant w} \right) =  \hfill \\
   = \frac{1}
{{C\;C'}}\left( {\frac{1}
{w}\left( {\int_{x = y}^{\,w} {x^{\,k} e^{\, - \,\lambda \,x} dx} } \right) + \frac{{w - y}}
{w}y^{\,k} e^{\, - \,\lambda \,y} } \right)dy \hfill \\ 
\end{gathered} 
$$
where $C'$ and thus $CC'$ shall be such as to normalize $p(y)$, i.e.:
$$
C\;C'\;:\quad \int_{y = 0}^{\,w} {p(y)dy}  = 1
$$
So, actually, it doesn't matter to normalize $p(x)$ at the beginning.
