Let an LP of the form:
$\max \ cx$
$s.t. \ Ax= b,$
$\ \ \ \ \ \ \ \ x\geq0$
Let the optimal solution of the dual LP problem be $w^*$.
If the $k^{th}$ equation of the original primal problem is multiplied by a constant $a>0$, what is the new optimal solution of the dual problem?
The dual for the original problem is :
$$\min p'b$$
$$p'A \geq c'$$
Let $a_i'$ denotes the $i^{th}$ row of $A$, where $i$ takes value from $1$ to $m$.
Let's rewrite the dual again:
$$\min \sum_{i \neq k}p_ib_i+p_kb_k$$
$$\sum_{i \neq k} p_ia_i'+p_ka_k' \geq c'$$
Now let's write down the dual upon multiplying the $k^{th}$ row by $a>0.$
Let $q$ denotes the new dual variable.
$$\min \sum_{i \neq k}q_ib_i+q_k(ab_k)$$
$$\sum_{i \neq k} q_ia_i'+q_k(aa_k') \geq c'$$
We know that multiplying a row by a non-zero constant doesn't change the objective value.
Hence given $w^*$, we should find a $q$ that satisfies the new dual. A natural choice is
$$q_i=\begin{cases} w_i^* & , i\neq k\\ \frac{w_i^*}{a} & ,i=k\end{cases}$$
Check that the proposed solution shares the same objective value and it is feasible.
Remark: We just require $a$ to be non-zero.