Is this correct: $Q_p^q(\lambda) = (\delta_{ij} + \lambda\delta_{ip}\delta_{jq})$
With $Q_i^j(\lambda)$ is the unit matrix with $\lambda$ at the i-th row and j-th column (author calls it type II unit matrix, but shouldn't matter).
Am I wrong or is $\delta_{ip}\delta_{jq}$ just $n$ at $p=q$ and $0$ otherwise and therefore $Q_p^q(\lambda) \neq (\delta_{ij} + \lambda\delta_{ip}\delta_{jq})$ because the right side is $\lambda n$ at p-th row, q-th column and $1$ diagonally? Am I missing something?
Yes you are right, your matrix matrix $Q_p^q(\lambda)$ is the identity matrix plus the term $\lambda$ at position $p,q$. For example for dimension $n=4$
$$ Q_2^3(\lambda) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & \lambda & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$
PS
Note that to be precise you should write $[Q_p^q(\lambda)]_{i,j} = \delta_{i,j} + \lambda \delta_{i,p} \delta_{j,q}$, as these are the components of the matrix.