Answer
See below
Work Step by Step
Let $B$ be the matrices obtained by adding column $i$ by column $j$ from $A$. We have $\det A=\det B$ because of a related property of P3 that is true for elementary olumn operations.
Since $\sum_ka_{ki}C_{kj}=0$ and $\det A=\sum_ka_{kj}C_{kj}$, cofactor expansion along column $j$ gives::
$\det A=\det B=$$\sum (a_{ki}+a_{kj})C_{kj}\\=\sum a_{ki}C_{kj}+\sum a_{kj}C_{kj}\\
=\det A+\sum a_{ki}C_{kj}$
