It can happen that $(\lambda+1)^2P-(\lambda a+d)(a+\lambda d)$ has multiple roots for any integer $\lambda$: namely, if $P=ad$, then
\begin{multline*}
(\lambda+1)^2P-(\lambda a+d)(a+\lambda d) \\
= (\lambda^2+1)(P-ad) + \lambda(2P-a^2-d^2)
= -\lambda (a-d)^2,
\end{multline*}
so that any root of $a-d$ is a multiple root of $(\lambda+1)^2P-(\lambda a+d)(a+\lambda d)$.

### A little beyond

One can actually classify completely the cases where $\lambda_n$ with the property in question can be found. Namely, from the identity
$$ (\lambda+1)^2P-(\lambda a+d)(a+\lambda d)
= (\lambda+1)^2(P-ad)-\lambda(a-d)^2 $$
it follows that if some root of $a-d$ is a multiple root of $P-ad$, then the polynomial in the left-hand side has multiple roots *for any integer $\lambda$*. On the other hand, if none of the roots of $a-d$ is a multiple root of $P-ad$, then there are only finitely many those $\lambda$ for which the polynomial has multiple roots; this follows from the general fact that if the polynomials $P$ an $Q$ do not have common multiple roots, then there are at most finitely many $\lambda$ for which $P+\lambda Q$ has a multiple root.