This is true.

I prefer to denote $q_i=p_i^{-1}$, $\alpha_i=a_ip_i$, $\beta_i=b_ip_i$, $A_i=\sum_{j=1}^i\alpha_j$, $B_i=\sum_{j=1}^i\beta_j$. Now we have to check that 
$$
\sum_i q_i\beta_iA_i^k\le C\sum_i q_i\alpha_iA_i^k+C\sum_i q_i\beta_iB_i^k
$$
This is linear in $q_i$, so we just need to check that 
$$
\sum_{i=1}^n\beta_iA_i^k\le C\sum_{i=1}^n\alpha_iA_i^k+C\sum_{i=1}^n\beta_iB_i^k
$$
for all $n\ge 1$.
But $xX^k$ is comparable with $X^{k+1}-(X-x)^{k+1}$ for $0\le x\le X$, so the right hand side is essentially $A_n^{k+1}+B_n^{k+1}$ and the left hand side is dominated by $(A_n+B_n)^{k+1}$. The rest should be clear.
  
Edit: To cover your second inequality, let's show that the "missing part"
$$
\sum_{i=1}^n p_i^k b_i\left(\sum_{j=i}^n a_j\right)^k\le
C\sum_{i=1}^n p_i^k a_i\left(\sum_{j=i}^n a_j\right)^k+
C\sum_{i=1}^n p_i^k b_i\left(\sum_{j=i}^n b_j\right)^k
$$
holds.
By now it shouldn't be surprising that it will suffice to check it for the sequence $p_i$ consisting of several zeroes followed by several ones, in which case it is just exactly the same story as before but written backwards (with summations starting with $n$ and going down).