= Solution
The hyperplane sections through $P$ are spanned by the three linear forms $x_1,x_2,x_0-x_3$. Their common zero on $X$ is precisely $P$, so away from $P$ the <linear system of divisors> is base-point-free and defines the <projection from a point on a smooth quadric>
$$
\boxed{\varphi([x_0:x_1:x_2:x_3])=[x_1:x_2:x_0-x_3].}
$$
In particular, the chart-ratio construction in part (a) makes this a <morphism of schemes> on $X\setminus\{P\}$, not merely a rational map there.
Write a target point as $[u:v:w]$. The corresponding line through $P$ consists of points
$$
[\lambda+\mu w:\mu u:\mu v:\lambda].
$$
Substituting in the quadric equation gives
$$
\mu\bigl((u-v)\lambda+uw\mu\bigr)=0.
$$
Removing $P$ means $\mu\ne0$, so set $\mu=1$. Over the <residue field> $K$ of the target point, its <scheme-theoretic fibre> is consequently
$$
\boxed{\operatorname{Spec}K[\lambda]/\bigl((u-v)\lambda+uw\bigr).}
$$
If $u\ne v$, this is one reduced point, with $\lambda=-uw/(u-v)$. If $u=v$ and $uw\ne0$, it is empty: the corresponding line meets the quadric only at the removed point, with intersection multiplicity two. If $u=v$ and $uw=0$, the line lies entirely on the quadric, and the <scheme-theoretic fibre> is $\mathbb A^1_K$, the line with $P$ removed.
The last case occurs at exactly $[0:0:1]$ and $[1:1:0]$. Their lines are respectively
$$
L_1=\{x_1=x_2=0\},\qquad L_2=\{x_0=x_3,\ x_1=x_2\},
$$
the two <rulings of a smooth quadric surface> through $P$. Thus the image consists of the complement of the line $u=v$, together with those two points. The line $u=v$ parametrizes directions in the tangent plane $x_1=x_2$ at $P$. The calculation describes the <scheme-theoretic fibres> over arbitrary target points and works in every characteristic.
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