\(5^3x+5^2y+5z+p=264\)

\(125x+25y+5z+p=264\)

\(125x+25y+5z=264-p\)

Since \(p\) is less than or equal to \(5\) and all the \(L.H.S\) elements are multiples of \(5\), to convert \(R.H.S\) into multiple of 5, let's assume \(p=4\)

\(125x+25y+5z=264–4\)

\(125x+25y+5z=260\)

Divide by \(5\)

\(25x+5y+z=52\)

Now, similarly to convert \(R.H.S\) into multiple of \(5\), let's assume that \(z=2\)

\(25x+5y= 52–2\)

\(25x+5y=50\)

\(5x+y=10\)

Now, similarly to convert \(R.H.S\) into multiple of \(5\), let's assume that \(y=5\)

\(5x=5\)

\(x=1\)

So, \(x+y+z+p= 1+5+2+4 = 6+6 =12\)

Answer C