= Solution
First consider the unlettered <projective frame> assertion. A list of $n+2$ points in $\mathbb{CP}^n$ is in general position when every $n+1$ of their representative vectors are <linearly independent>, equivalently no $n+1$ of the points lie in a projective hyperplane. Choose representatives $v_1,\ldots,v_{n+1}$ of the first $n+1$ points as a <basis>. Write $v_{n+2}=\sum c_iv_i$. Each $c_i$ is nonzero, since otherwise $v_{n+2}$ and the other $n$ basis vectors would be dependent. Rescaling the representatives makes $v_{n+2}=v_1+\cdots+v_{n+1}$.
Do the same for the second frame with representatives $w_i$. The <invertible linear map> $v_i\mapsto w_i$ induces a <projective transformation> carrying all $n+2$ points as required. A transformation fixing a normalized frame is represented by a diagonal matrix, since it fixes the first $n+1$ points. Fixing the sum point forces every diagonal entry to coincide, so the matrix is scalar and the <projective transformation> is the identity. This proves \b[existence and uniqueness].
Return to the four chosen points on the two lines. Since the six points are distinct, at most one is $K\cap L$; put its index third if necessary. No three of $p_1,p_2,q_1,q_2$ are collinear: every triple contains two on one line and a point on the other line away from the intersection. Thus they form a <projective frame> in $\mathbb{CP}^2$. The four target points also have no collinear triple, as their three-by-three <determinants> are nonzero. The frame result gives \b[a unique <projective transformation> with the prescribed four images].
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