Proving RTln($a_i$) equal to non PV work done by system,where $a_i$ is activity
We know that chemical potential is defined as
  ${\mu}_i={\mu}_i^{standard}+RT\ln(a_i)$.


Be a bit careful with your wording. This equation is not part of the definition of the chemical potential but rather it is derived from it.
The derivation of this equation can be found in this answer of mine.
The definition of the chemical potential is

\begin{equation}
 \mu_{i} = \left( \frac{\partial G}{\partial n_{i}} \right)_{p,T} \ ,
\end{equation}

where $n_{i}$ is the amount of the $i^{\text{th}}$ component in the system.


  We know that Gibbs free energy of a system at constant temperature and
  pressure also represents Non-PV work done by system.


Again this is not quite correct: It is the change in Gibbs free energy of a system during some process, say a chemical reaction or a phase transition, at constant temperature and pressure that can be identified with non-PV work done by the system.

Now considering a general chemical reaction

\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C &lt;=&gt; \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}

the differential change in Gibbs free energy is given by

\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \mu_{i} \text{d} n_{i}
\end{equation}

or introducing the extend of reaction, $\text{d} \xi = \frac{\text{d} n_{i}}{\nu_{i}}$

\begin{equation}
 \text{d} G = -S \text{d} T + V \text{d} p + \sum_{i} \nu_{i} \mu_{i} \text{d} \xi \ .
\end{equation}

From this it follows that the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$) at constant pressure and temperature is given by

\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ .
\end{equation}

Here you can see that the non-PV work done by the system originates from the differences in the chemical potentials of the reactants.
Substituting the equation for the chemical potential

\begin{equation}
 \mu_{i} = \mu_{i}^{0} + RT \ln a_{i} 
\end{equation}

the equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, leads to

\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}

where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced.
This is the relation between the non-PV work, represented by $\Delta G$, and the activity terms. Admittedly, it is not a very graphic explanation but I fear that it simply is a pretty mathematical matter.