 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 $v_1,\dots ,v_{n+1}$ of the first $n+1$ points as a basis. Write $v_{n+2}=\sum c_i{v}_i$. 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 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 a unique projective transformation with the prescribed four images.