You only have to use \tkzGetPoint{C} instead of \tkzGetPoint(C). Syntax change.
I also added (drawing) the points and lines with \tkzDrawPoints(C,X_1) and \tkzDrawLine(A,B) to visualize the solution. \tkzLabelPoints(A,B) plots the label.
Define a orthogonal line to a other line (A,B) going through a Ponit P:
Therefore you can use \tkzDefLine[orthogonal=through P](B,A), the second point of the line you can get with \tkzGetPoint{X_1}(first point is P) defined as X_1.
Compute the intersection of two line:
\tkzInterLL(P,X_1)(A,B): computes the intersection of the two lines (P,X_1) and (A,B). With \tkzGetPoint{C} you get the intersection point, saved as C.
Solution:

MWE:
\documentclass{article}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}
\tkzDefPoint(0,0){A}
\tkzDefPoint(11,0){B}
\tkzDefPoint(10,10){P}
\tkzDrawPoints(A,B,P)\tkzLabelPoints(A,B,P)
%
\tkzDefLine[orthogonal=through P](B,A)
\tkzGetPoint{X_1}
%
\tkzInterLL(P,X_1)(A,B)
\tkzGetPoint{C}
\tkzDrawPoints(C,X_1)\tkzLabelPoints(C,X_1)
%
\tkzDrawLine(A,B)
\tkzDrawLine(P,X_1)
\end{tikzpicture}
\end{document}