The answer is $2k+1$. For the construction part, consider regular $(2k+1)$-gon $V_1V_2...V_{2k+1}$ centered at $P$. Use the $2k+1$ lines $V_iV_{i+1}$ where $i=1,2,...,2k+1$ (indices taken in modulo $2k+1$.) Note that if we extend the sides of the polygon, we can divide the whole pictures into $k$ layers of edge-disjoint (possibly concave) polygons, all contains $P$ in its interior. This means any ray starting at $P$ must intersect these layers $k$ times, each with different lines. For the prove, suppose it can be done by drawing only $2k$ lines. Consider a line $\ell$ pass through $P$ parallel to one of those lines. This line $\ell$ divided into two rays. Also, $\ell$ intersect with those $2k$ lines at most $2k-1$ times. Hence, there's a ray starting from $P$ that intersects at most $k-1$ times, contradiction.