$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\epi}{\operatorname{epi}}$Yes, $p$ is Lipschitz. 

Indeed, 
\begin{equation*}
	p(x)=x+f(x)u
\end{equation*}
for $x\in C$, where 
\begin{equation*}
	f(x):=\inf E_x,\quad E_x:=\{t\ge0\colon x+tu\notin C\}. 
\end{equation*}
Let us use the [Rockafellar, p. 62][1] notation $0^+C$ for $C^{as}$ and then $(0^+C)^\circ$ for the interior of $0^+C$. 

So, it suffices to show that the function $f$ is Lipschitz on $C$. 

For each $x\in C$, we have 
\begin{equation*}
	f(x)=\infty\iff E_x=\emptyset\iff (\forall t\ge0\ x+tu\in C)\iff u\in0^+C; 
\end{equation*}
the latter $\iff$ follows by [Rockafellar, Theorem 8.3, p. 63][1]. So, in view of the condition $u\notin0^+C$, we see that 

\begin{equation}
	\text{$0\le f<\infty$ on $C$. } \tag{0}\label{0}
\end{equation}

Without loss of generality (wlog), the vector $u$ is of unit length. 
Since 
\begin{equation*}
	v:=-u\in(0^+C)^\circ,
\end{equation*}
there is some real $\ep>0$ that 
\begin{equation*}
	v+\ep B\subseteq 0^+C, \tag{1}\label{1}
\end{equation*}
where $B$ is the closed unit ball centered at the origin. 

It is enough to show that the restriction of $f$ to the intersection of $C$ with any 2-D affine plane parallel to the vector $u$ is $L$-Lipschtiz, with the same Lipschitz constant 
\begin{equation*}
	L:=1+l,\quad l:=1/\ep \tag{2}\label{2}
\end{equation*}
for all such 2-D affine planes. 

Therefore, wlog the dimension $n$ is $2$. Also then, wlog $-u=v=(0,1)$. It then follows that 
\begin{equation*}
	f(a,b):=f((a,b))=b-g(a) \tag{3}\label{3}
\end{equation*}
for $(a,b)\in C\subseteq\R^2$, where 
\begin{equation*}
	g(a):=\inf\{y\in\R\colon(a,y)\in C\}
\end{equation*}
for all real $a$, so that $C$ is the epigraph $\epi(g):=\{(a,y)\in\R^2\colon y\ge g(a)\}$ of the (necessarily convex) function $g$. Here, the standard convention $\inf\emptyset=\infty$ is used (even though we will see in a moment that in our case in fact we have $g(a)<\infty$ for all real $a$. 

By \eqref{1} and \eqref{2}, $\epi(g)$ contains $\epi(h)$ for a function $h$ given by a formula of the form $h(a)=B+l|a-A|$ for some real $A,B$ and all real $a$. Then $g\le h$; in particular, $g<\infty$ on $\R$. Also, by \eqref{3} and \eqref{0}, $g(a)>-\infty$ for some real $a$. So, for the convex function $g$ we have $g>-\infty$ on $\R$. So, $-\infty<g<\infty$ on $\R$. 

Now it follows from $g\le h$ that the limit $g'(\infty)$ of the (say) right derivative $g'(a)$ of the (finite convex) function $g$ as $a\to\infty$ is $\le h'(\infty)=l$. Similarly, the limit $g'(-\infty)$ of the right derivative $g'(a)$ of the convex function $g$ as $a\to-\infty$ is $\ge h'(-\infty)=-l$. Since the right derivative $g$ of the convex function $g$ is nondecreasing, we see that $-l\le g'\le l$. So, $g$ is $l$-Lipschitz. 

Thus, by \eqref{3} and \eqref{2}, $f$ is indeed $L$-Lipschtiz. $\quad\Box$ 

[1]:https://www.google.com/books/edition/Convex_Analysis/1TiOka9bx3sC?hl=en&gbpv=1