Assume we have the matrix product: $$A=PBD$$

where $P$ is a projection matrix (i.e., $P=P^2$, $P=P^\top$, and $\|P\|_2=1$), $B$ is a matrix whose infinite norm is equal to one ($\|B\|_\infty=1$), and $D$ is a diagonal matrix whose $\ell_2$-norm is less than one ($\|D\|_2<1$).

Is it correct to say that the spectral radius of $A$ is less than 1 ($\rho(A)< 1$). If yes, how to prove it?

Yes. Denote the entrywise absolute value of a complex matrix $X$ by $|X|$. In general, if $|X|\le Y$ entrywise, then $\rho(X)\le\rho(|X|)\le\rho(Y)$.
Your $P$ is an orthogonal projection. Therefore the moduli of its entries are bounded above by $1$. Hence $|A|\le|B||D|$ and $\rho(A)\le\rho(|B||D|)$. So, it suffices to prove that the latter quantity is strictly smaller than $1$.
Suppose $(\lambda,v)$ is an eigenpair of $|B||D|$. Clearly, $\|\,|B|\,\|_\infty=\|B\|_\infty=1$. As $D$ is diagonal, $\|\,|D|\,\|_\infty=\|D\|_2<1$. Therefore $|\lambda|\|v\|_\infty=\|\,|B|\,|D|\,v\|_\infty\le\|\,|B|\,\|_\infty\|\,|D|\,\|_\infty\|v\|_\infty<\|v\|_\infty$. Thus $\rho(|B||D|)<1$. $\square$
Remark. Your conjecture is true because $D$ is diagonal. Otherwise it is false in general. Here is a counterexample. Let $B=\pmatrix{-\frac12&-\frac12\\ 0&1}$. Its largest singular value is $\sigma_1=\frac14(3+\sqrt{5})>1$. Let $B=USV^T$ be its singular value decomposition. Set $P=U\pmatrix{1\\ &0}U^T$ and $D=rVU^T$ for some $r$ that is smaller than but sufficiently close to $1$. Then $A=PBD=r\sigma_1U\pmatrix{1\\ &0}U^T$, whose spectral radius is $r\sigma_1>1$.
• Thank you for your useful response. One additional question. If I have a block diagonal matrix $D$ instead of a diagonal one where each block $D_i$ is symmetric and $\|D_i\|_2<1$. Furthermore, I know that $D$ is diagonally dominant. Is it possible to prove that $\rho(A)<1$? – user293017 Mar 31 '16 at 9:14
• @user293017 The $D$ in the counterexample given in my answer is symmetric and diagonally dominant. Numerically $D\approx r\pmatrix{-0.9487&-0.3162\\ -0.3162&0.9487}$. So, using the $P,B$ and $D$ in the remark, $\pmatrix{P\\ &P}\pmatrix{B\\ &B}\pmatrix{D\\ &D}$ gives you a counterexample to your new conjecture.. – user1551 Mar 31 '16 at 19:17