 Solution

Take $D=\{y^{2}z=x^3\}$, $L=\{z=0\}$ and $C=3L$. This projective cuspidal cubic is irreducible and meets $L$ only at $p=[0:1:0]$. At $p$, the derivative with respect to $z$ is $y^2=1$, so $D$ is smooth. In the affine chart $y=1$, it is $z=x^3$, and $L$ is $z=0$, so $I_p(L,D)=3$. Multiplicity is linear in the divisor $C=3L$, giving