The given equation is $(2 p+1) x^{2}-(7 p+2) x+(7 p-3)=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=2 p+1, b=-(7 p+2)$ and $c=7 p-3$.

$\begin{array}{l}
\therefore D=b^{2}-4 a c \\
=-[-(7 p+2)]^{2}-4 \times(2 p+1) \times(7 p-3) \\
=\left(49 p^{2}+28 p+4\right)-4\left(14 p^{2}+p-3\right) \\
=49 p^{2}+28 p+4-56 p^{2}-4 p+12 \\
=-7 p^{2}+24 p+16
\end{array}$

The given equation will have real and equal roots if $D=0$.

$\begin{array}{l}
\therefore-7 p^{2}+24 p+16=0 \\
\Rightarrow 7 p^{2}-24 p-16=0 \\
\Rightarrow 7 p^{2}-28 p+4 p-16=0 \\
\Rightarrow 7 p(p-4)+4(p-4)=0 \\
\Rightarrow(p-4)(7 p+4)=0 \\
\Rightarrow p-4=0 \text { or } 7 p+4=0
\end{array}$

$\Rightarrow p=4$ or $p=-\frac{4}{7}$

Hence, 4 and $-\frac{4}{7}$ are the required values of $p$.